cool thank you so much
#algos-and-data-structs
jaunty loom
jaunty loom
what is the time complexity of ```py
def myFunc(str):
inputOX = str.lower()
ohh = inputOX.split('x')
exx = inputOX.split('o')
return len(ohh)==len(exx)
trim fiber
what is the time complexity of ```py
def myFunc(str):
inputOX = str.lower()
...
O(n)? .lower is O(n), .split seems to be O(n) (I am not sure), len(...) is O(n)
jaunty loom
`O(n)`? `.lower` is `O(n)`, `.split` seems to be `O(n)` (I am not sure), `len(.....
split seems like its O(n*n)
since if every item in the string is 'x' the it would have to split every component of the list
i dunno it it seems weird
main flower
split probably could be done in O(n) If I'm not wrong
which would make the whole function O(n)
jaunty loom
oh
trim fiber
!e
def split(string, delimiter):
output = [""]
for character in string:
if character != delimiter:
output[-1] += character
else:
output.append("")
return output
print(split("ab1cb1de", "1"))
halcyon plankBOT
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
['ab', 'cb', 'de']
lean dome
split is O(n). It creates at most 1 element in the resulting list per character in the original string, so that's 1 thing done N times, O(n), not N things done N times, O(n**2)
fervent saddle
split should always be O(n) if itβs splitting by a single character I think
jaunty loom
what is the time complexity of
array = [1, 2, 3, 4]
x = array[:]```O(N)
jaunty loom
so wouldn't split have to do that for each split?
fervent saddle
I remember reading that split is O(n*m) in the worst case, with n being the length of original string and m being the length of the substring. So if the substring is just one character, itβs O(n)
lean dome
That's right.
jaunty loom
hmm
lean dome
so wouldn't split have to do that for each split?
split() returns a list, not a list of lists.
fervent saddle
And there are algorithms that could make it O(n) or better always, but theyβre more expensive, so they just use a cheaper O(n*m) algorithm
jaunty loom
oh yeah that makes sense
assuming the list is
string = "xxxxxxxxxxx"
string.split('x')
``` wouldn't it have to
```py
string = 'xxxxxxx'
list = []
index = 0 # index of the x
for i in number of 'x' in the string:
list.append(string[index:(next_index_of_x)])
index = next_index_ofx```
is this not how the split is worked out?(not valid python code sorry)
lean dome
Yep, that's basically right.
jaunty loom
then from the line
for i in number of 'x' in string:
it would be O(n)
and in the next line
list.append(string[index:(next_index_of_x)])
it would also be O(n) right?
which would lead to
O(n) * O(n)
= O(n*n)?
lean dome
list.append is O(1), not O(n)
main flower
Well, technically yes but you could have done that the other way in O(n)
main flower
`list.append` is O(1), not O(n)
he probably meant slicing
jaunty loom
`list.append` is O(1), not O(n)
the string[index:...] part
lean dome
Slicing the string to extract a substring is O(m), where m is the length of the substring to be extracted, not O(n), where n is the length of the original string
main flower
s = "abcxdefxaaa"
def split_str(s, delimeter):
out = []
temp_str = ""
for char in s:
if char == delimeter:
out.append(temp_str)
temp_str = ""
else:
temp_str += char
if temp_str != "": out.append(temp_str)
return out
split_str(s, "x")
This could work
and here it's just O(n) cuz you traverse the string only once
(unless there are some errors xd)
main flower
well, I keep the string and then when I find delimeter I just append string to the list and reset it
lean dome
The better way to view this is that the algorithm needs to read every character of the string being split twice (assuming single character delimiters): once to see if it's a delimiter or not, and once to copy it into a string in the resulting list if it's not. That's O(2*n). Which is O(n).
Both adding a new string to the list of strings being returned, and adding a new character onto one of those strings, are amortized O(1) operations, we can ignore them.
jaunty loom
Both adding a new string to the list of strings being returned, and adding a new...
but in the first algorithm doesn't it also create new lists during the slicing of the string which would be O(m)?
lean dome
Yes, but it still only visits each character twice. Once during the for loop to see if it's a delimiter or not, and once as part of a slice if it's not.
main flower
yeah
jaunty loom
oh
lean dome
You're confusing yourself by adding an M in where you don't need one. You can formulate it as O(n*m) if you really want to, but n*m is at most n*2.
jaunty loom
im a little bit confused about the slicing part, but seems like it would still be O(1) since its basically the same string but just cut down
anyways, i gotta go to the school, thnx for the explanation @lean dome and @main flower
main flower
π
lean dome
im a little bit confused about the slicing part, but seems like it would still b...
Extracting a slice is O(len(slice)). But we know there's an upper bound on how many slices we extract (n) and on the length of each of those slices at the upper bound (0, if every character was a delimiter and there are no non-delimiter characters to copy). That's creating N empty strings, which is O(n) total across all iterations, because creating an empty string is O(1).
And conversely, there's a lower bound on how many slices we extract - it's 1, in the case where the string contained no delimiters, in which case we extract an N character substring 1 time, which is also O(n) total across all iterations.
The trick here is that N and M aren't independent variables, they're inversely proportional. If the loop fires more times, it does less work per iteration, and vice versa.
And the total amount of work it does across all iterations is O(n), no matter how many iterations there are.
halcyon sequoia
Could someone help me with this task?
Construct recursive algorithms in the form of programs.
(3,5,3,-1,-3,-1,3,5...)
(1,5;1;0,5;-0,5;-2;-4,5; -8,5)
(-3,1,-4, -5,19,-96,-1825,...)
manic swift
if i add another name, then where will the rear pointer go? its a circular queue btw
tropic nacelle
any idea how can I set a list as an entire file
like
file is [item, item , item] (5 million items)
lean dome
if i add another name, then where will the rear pointer go? its a circular queue...
Given that diagram, and the most common implementation of a circular queue, you cannot add another item. The queue is full.
trim fiber
file is [item, item , item] *(5 million items)*
file.readline()?
lean dome
Given that diagram, and the most common implementation of a circular queue, you ...
There are possibly implementations that make use of every cell of the array, but the naive implementation you're likely to be learning first has one element of overhead.
manic swift
Given that diagram, and the most common implementation of a circular queue, you ...
ahh okay but i thought circular queues make use of all the memory spaces?
lean dome
That's because, if it were pushed back one further, it would be equal to the front pointer - and that's the representation you've chosen for an empty array, so it can't also be the representation for a full array
lean dome
ahh okay but i thought circular queues make use of all the memory spaces?
They can. The naive implementation does not.
manic swift
ah okay so it can, but not in the way that the diagram above is shown?
also, can priority queues be both linear and circular?
agile sundial
circular priority queue doesn't make sense, where is the highest priority
lean dome
A priority queue isn't even really a queue... It doesn't have any order. They can't be circular, because there's no order at all, and so no beginning or end.
lean dome
ah okay so it can, but not in the way that the diagram above is shown?
In some languages, with some constraints, it is possible to fully utilize the buffer. The trick I know of for implementing that relies on either arbitrary precision integers like Python has, or defined behavior for unsigned overflow like C has (in which case it only works for buffers whose size is a power of two)
manic swift
A priority queue isn't even really a queue... It doesn't have any order. They ca...
so are they just.. a normal linear queue
agile sundial
usually you implement one with a heap
that way higher priority things are popped first
lean dome
They're not linear, because they don't have an order, just a concept of "current highest priority item"
agile sundial
pick up a textbook or online course and get to self studying lol
lean dome
"Priority queue" is really a misnomer, because there's no "waiting in line" like there is for a regular queue. If a high priority item shows up, it's immediately moved straight to the top to be processed next. Though items of equal priority may or may not preserve their relative order.
agile sundial
what should it be named instead?
manic swift
one more thing, can you explain 'checksums' please?
lean dome
what should it be named instead?
"Priority heap" would be more accurate, I suppose. The problem is that "stack" and "queue" are both named after metaphors that don't apply to priority queues.
agile sundial
huh, are there (good) implementations that don't use heaps? if there were you'd start calling things that aren't heaps heaps
lean dome
I'm not sure. "Priority bag"? "Priority bunch"?
lean dome
one more thing, can you explain 'checksums' please?
They're a specific purpose of hashing. They're hash algorithms designed to make it very likely that minor corruptions of the input data lead to very different hash codes, so that, if applying the same hash algorithm to the data you received matches the hash value recorded by the sender, you can be reasonably confident that the data was transferred successfully.
manic swift
sorry, so its an algorithm used in data transmission, to check if the data received is legit and hasn't been like tampered with? and the end user runs the same algorithm and if the checksum digit matches then they know there weren't any errors in the transmission. is that correct?
silk arrow
can someone help me?
spring jasper
Not unless you explain what you need help with
remote garnet
how do i make it so that if a valid operator isnt inputted, it goes back to the beginning?
n1 = int(input("Enter number: "))
n2 = int(input("Enter another number: "))
op = str(input("Enter an operator (+ - / *): "))
if (op == "+"):
print(n1+n2)
elif (op == "-"):
print(n1-n2)
elif (op == "*"):
print(n1*n2)
elif (op == "/"):
print(n1/n2)
else:
print("Enter a valid operator: ")
also if this is the wrong channel can some redirect me to the right one
jaunty loom
how do i make it so that if a valid operator isnt inputted, it goes back to the ...
put it in a while loop i guess
fiery cosmos
how do i make it so that if a valid operator isnt inputted, it goes back to the ...
loop = True
while loop:
"""
your code
"""
loop = input('Do you want me to stop(yes/no): ')
if loop == 'yes':
break
else:
continue
glossy crypt
Hello! I have to solve these recurrences using the master theorem. I was wondering about the second one... Can I simply multiply the recurrence by 2 and then solve, such that 64 T((n/4)*2) + n^4. Thaanks!
elfin citrus
Could someone help me answer this question? My mind has gone blank and I cannot figure out how to answer this question of Data Encoding with Python.
wet urchin
Good morning everyone
I need your help, I'm trying to understand an exercise but I can't understand, it's about enqueue with linked list.
wet urchin
Someone now how to copy from this webpage? https://paste.pythondiscord.com/
trim fiber
Someone now how to copy from this webpage? https://paste.pythondiscord.com/
You have Save option and there is a unique link in your browser bar
wet urchin
wet urchin
I need to know, how enQueue works
why he puts self.lastNode = self.front? It should be =self.rear I think
trim fiber
why he puts self.lastNode = self.front? It should be =self.rear I think
In your paste there is self.lastNode = self.read 
def enQueue(self, data):
self.lastNode = self.rear
self.rear = Node(data)
if self.lastNode:
self.lastNode.setNext(self.rear)
if self.front is None:
self.front = self.rear
self.size += 1
First of all you don't need functions like setLast, getLast and so on in Node class
It can be done just by node.last = newlast, there is no need to make it by node.setLast(newlast)
wet urchin
Why you think that the book puts that in that way?
trim fiber
Why you think that the book puts that in that way?
Quick look at the code and implementation of QueueLinkedListsCircular.enQueue looks okay for me
def enQueue(self, data):
self.lastNode = self.rear # create temporary variable with last rear
self.rear = Node(data) # set new node as new rear
if self.lastNode: # if last rear is not None then set its new node as new rear
self.lastNode.setNext(self.rear)
if self.front is None: # if front is none then set front as rear, it means that self.lastNode is also None
self.front = self.rear
self.size += 1
Thanks a lot
Maybe may you put a comments for the deQueue method please?
I don't understand what is the function of result
trim fiber
Maybe may you put a comments for the deQueue method please?
def deQueue(self):
if self.front is None: # if queue is empty then raise exception
print('Sorry, the queue is empty..!')
raise IndexError
# it means that self.front is not None
result = self.front.getData() # get data of the front node (this data is set in constructor)
self.front = self.front.next # set front as next node
self.size -= 1
return result;
Thanks a lot MK
frank osprey
Hii! π could someone please help me with the following problem: Describe how to implement a dictionary that has an average lookUp, insert, and delete time of O(1), but uses an array of no more than 2 times the maximum number of elements that will be stored in the dictionary. So, if I have a student node class and the keys would be represented by their last 4 digits of the SSN, and there are 20 students, I would create a dictionary of size 40 and technically use the keys as index values in order to keep the search time as constant as possible, but in this case I can sacrifice a bit the best search time for a smaller array size. Would I have to use something like hashtables? (i am not sure how they work, I just looked them up).
elfin citrus
Hey guys, there is a better way to write this..... right?
test1_in = [28, 2, 3, -3, -2, 1, 2, 35, -1, 0, 0, -1]
test0 = test1_in[0]
test1 = test0 + test1_in[1]
test2 = test1 + test1_in[2]
test3 = test2 + test1_in[3]
test4 = test3 + test1_in[4]
test5 = test4 + test1_in[5]
test6 = test5 + test1_in[6]
test7 = test6 + test1_in[7]
test8 = test7 + test1_in[8]
test9 = test8 + test1_in[9]
test10 = test9 + test1_in[10]
test11 = test10 + test1_in[11]
print(str(test0) + ' ' + str(test1) + ' ' + str(test2) + ' ' + str(test3) + ' ' + str(test4) + ' ' + str(test5) + ' ' + str(test6) + ' ' + str(test7) + ' ' + str(test8) + ' ' + str(test9) + ' ' + str(test10) + ' ' + str(test11))
old rover
test1_in = [28, 2, 3, -3, -2, 1, 2, 35, -1, 0, 0, -1]
test0 = test1_in[0]
test1 = test0 + test1_in[1]
test2 = test1 + test1_in[2]
test3 = test2 + test1_in[3]
test4 = test3 + test1_in[4]
test5 = test4 + test1_in[5]
test6 = test5 + test1_in[6]
test7 = test6 + test1_in[7]
test8 = test7 + test1_in[8]
test9 = test8 + test1_in[9]
test10 = test9 + test1_in[10]
test11 = test10 + test1_in[11]
seems like it could be just done with sum(test1_in)
saves you a lot of code
or you could do it with a for loop
wait
elfin citrus
but I dont want the sum
old rover
hm
elfin citrus
or you could do it with a for loop
Yea was thinking of doing that, but don't know how
frank osprey
Yea was thinking of doing that, but don't know how
could you use recursion?
elfin citrus
I can use any method but the output has to be similar or same to the one I wrote above
native oar
I have a graph, and i have implemented a BFS to find the shortest path, in terms of edges, for the graph, given two vertices, now the order of growth to find the the shortest path would be O(V + E) i think, but how would i test this empirically using timeit? Would i use the graph i currently have, and pick two random vertices and measure time and do some reps? Like usually (for other stuff) i would have plotted a running time against problem size graph, but here i am not sure how i would do it. Or would i need to create a multiple new graphs just to test this?
frank osprey
I can use any method but the output has to be similar or same to the one I wrote...
well then I can't think from top of my head how to specifically do it (i came here for help too lol), but you should have let's say a Fib() method with a base case for returning test1_in[0] then in the recursive case, you would have something like previous element + next element, similar to a fibonacci series
dense furnace
Hey everyone, I'm running into a problem that should be simple, but is perplexing me. I'm trying to plot a 3D surface, where I have a 2D array of radii, and each axis represents theta and phi in spherical coordinates. I'm using pyqtgraph and OpenGL to render the 3D images.
I can calculate all x, y, z locations for each r, theta, phi, and stack them into an Nx3 array, however converting that to a surface is difficult because they are not evenly distributed in Cartesian coordinates (they are evenly distributed in spherical coords). From digging around, it looks like I should create a mesh out of these points, however I am struggling with how to do that. Can anyone help me out here?
wet urchin
.
true fossil
this isnt an alg per-cay and it isnt python specific but lets say i had 2 linear equations in slope-intercept form (y=mx+b). what would be the logic behind finding the point of intersection (if there is one). would i have to check every point on both lines??? that seems like it would take a while... i could check to see if the slope is the same to see if they are parallel or the same line(s)
agile sundial
you can calculate a point of intersection in constant time
since the y will be the same at the intersection, just set the two equations equal
m1 * x + b1 = m2 * x + b2
solve for x
true fossil
right but besides setting the 2 equations equal to each other, how would i code that
ik how to do it irl
but it depends on the situation what you are going to do
unreal mason
right but besides setting the 2 equations equal to each other, how would i code ...
I found something that might help you:
https://stackabuse.com/solving-systems-of-linear-equations-with-pythons-numpy/
Specifically look for the Solver() method in this page.
This link has the explanation too:
https://apmonitor.com/che263/index.php/Main/PythonSolveEquations
sacred aspen
hello
sacred aspen
how are you
trim fiber
how are you
Fine, thanks. This channel is dedicated to talk about algorithms and data structures.
If you want to talk about offtopic conversions here you have dedicated channels:
Feel free to join to one of those
cerulean river
Hey all, I have just solved this problem on Leetcode Problem 387: https://leetcode.com/problems/first-unique-character-in-a-string/ with the following code
class Solution:
def firstUniqChar(self, s: str) -> int:
d = dict()
index = -1
for i in s:
d[i] = d.get(i,0) + 1
for i,elem in enumerate(s):
if d[elem] == 1:
return i
return index
Although this solution works, and its complexities are as follows Runtime: O(N), Space: O(N) however whenever I run this on Leetcode it seems to have a horrible actual runtime. Below are the runtimes from my last 5 submissions.
spring jasper
Theyre accepted, whats the problem
cerulean river
I have also found other solutions on the discuss section and noticed when the Counter module is used, they typically have much better runtimes such as this code
class Solution:
def firstUniqChar(self, s: str) -> int:
from collections import Counter
d = Counter(s)
for i in d:
if d[i] == 1:
return s.index(i)
else:
return -1
This person has gotten
cerulean river
Theyre accepted, whats the problem
I was wondering why the Counter module yields much better results in actual runtimes. I was wondering if someone can explain to me why that is. From my understanding I am doing the same thing just using a dictionary but if this is a misunderstanding please so point it out. I am currently looking online for an explanation but figured I can ask here where I can ask about my code.
Wouldn't the counter be iterating s anyway so actual iterations should be the exact same just less code.
Just submitted this code myself and it is still performing better than my initial solution.
spring jasper
Hmm yea thats weird
Counter(s) should be O(len(s))
Oh, but
They have two loops, one over s and one over len(Counter)
You have two loops over s
Maybe thats why their runtime is shorter i would guess
cerulean river
but wouldn't Counter implicitly add a loop on s for the counting?
cerulean river
what is profiling in this context?
spring jasper
Basically checking whats taking long in your code
Anyway both code samples have a loop over s for counting
cerulean river
Oh I see
I also suspect that the return s.index(i) portion in the faster code would also be another loop, unless Counter is storing index somehow in the initial count.
spring jasper
It shouldn't, counter's items are the element and its count in the iterable
cerulean river
So you mean that if s = 'aabb' Counter(s) would produce
a: 2
b: 2
So kind of like a dictionary?
trim fiber
So you mean that if `s = 'aabb'` `Counter(s)` would produce
```
a: 2
b: 2
```
S...
!e
from collections import Counter
string = "aabb"
output = Counter(string)
print(output)
print(output["a"])
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
001 | Counter({'a': 2, 'b': 2})
002 | 2
Thank you!
trim fiber
Thank you!
I like to check how it works π
halcyon plankBOT
:incoming_envelope: :ok_hand: applied mute to @vernal glen until 2021-04-13 22:11 (9 minutes and 59 seconds) (reason: attachments rule: sent 8 attachments in 10s).
spring jasper
<@&267629731250176001>
torpid crown
!pban 583069436387524610 Spam
spring jasper
oops, false alarm i guess
halcyon plankBOT
:incoming_envelope: :ok_hand: applied purge ban to @vernal glen permanently.
cerulean river
Still not getting why using Counter to count is roughly 3x faster than using the dictionary. But I am guessing this is meaningless in an interview setting since they care about the algorithm rather than microoptimzations available in the language? Can someone please comment on this?
spring jasper
Anyway, the source to counter is here
https://github.com/python/cpython/blob/5ce227f3a767e6e44e7c41e0c845a83cf7777ca4/Lib/collections/__init__.py#L530
You can check exactly what it does
cerulean river
Anyway, the source to counter is here
https://github.com/python/cpython/blob/5ce...
Thank you for the link and trying to help me understand this.
spring jasper
You can see that theres a loop over the iterable to populate the counter object
And its very very close to what you wrote
cerulean river
I see, thank you!
trim fiber
Still not getting why using `Counter` to count is roughly 3x faster than using t...
Maybe the problem is with making Python objects - it's very costly
spring jasper
Why would dict() be so much slower than subclassing dict and using super() tho
cerulean river
Could it be my use of enumerate?
trim fiber
!e
from collections import Counter
from random import choice
from string import ascii_letters
from time import time
string = "".join([choice(ascii_letters) for _ in range(2 ** 16)])
start = time()
Counter(string)
end = time()
print("counter", end - start)
start = time()
counter = {}
for character in string:
if character not in counter:
counter[character] = 1
else:
counter[character] += 1
end = time()
print("dict", end - start)
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
001 | counter 0.00568389892578125
002 | dict 0.01647639274597168
There is significant difference
unique pecan
Hi
r = requests.get("https://www.xxx.net/api/get-engines?username=xxx&password=xxx")
returns
halcyon plankBOT
Hey @unique pecan!
unique pecan
{"file_engines":[{"id":"1","full_name":"Adaware Antivirus 12","engine":"adaware","version":"","update":""},{"id":"2","full_name":"AhnLab V3 Internet Security","engine":"ahnlab","version":"","update":""},{"id":"3","full_name":"Alyac Internet Security","engine":"alyac","version":"","update":""},{"id":"4","full_name":"Avast Internet Security","engine":"avast","version":"","update":""},{"id":"5","full_name":"AVG AntiVirus","engine":"avg","version":"","update":""},{"id":"6","full_name":"Avira Antivirus 2018","engine":"avira","version":"","update":""},{"id":"7","full_name":"Bitdefender Total Security 2018","engine":"bitdef","version":"","update":""},{"id":"8","full_name":"BullGuard Antivirus","engine":"bullguard","version":"","update":""},{"id":"9","full_name":"ClamAV","engine":"clam","version":"","update":""},{"id":"10","full_name":"Comodo Antivirus","engine":"comodo","version":"","update":""},{"id":"11","full_name":"Dr.Web Security Space 11","engine":"drweb","version":"","update":""},{"id":"12","full_name":"Emsisoft Anti-Malware","engine":"emsisoft","version":"","update":""},{"id":"13","full_name":"ESET NOD32 Antivirus","engine":"nod32","version":"","update":""},{"id":"14","full_name":"FortiClient Antivirus","engine":"forti","version":"","update":""},{"id":"15","full_name":"F-Secure SAFE","engine":"fsecure","version":"","update":""},{"id":"16","full_name":"IKARUS anti.virus","engine":"ikarus","version":"","update":""},{" AND MORE
how do I parse certain pieces of this data?
with requests
trim fiber
```{"file_engines":[{"id":"1","full_name":"Adaware Antivirus 12","engine":"adawa...
Have you tried result.json()? I don't know how do you want to parse it
unique pecan
Have you tried `result.json()`? I don't know how do you want to parse it
<bound method Response.json of <Response [200]>>
<bound method Response.json of <Response [200]>>
just returns that
trim fiber
<bound method Response.json of <Response [200]>>
<bound method Response.json of ...
Not result.json but result.json() - remember about ()
unique pecan
Ohhh let me try thx
unique pecan
Not `result.json` but `result.json()` - remember about `()`
returns same thing with spaces
There should be a way to parse the data no? how does everyone else do it?
trim fiber
There should be a way to parse the data no? how does everyone else do it?
Show result.headers['content-type'] output
unique pecan
Show `result.headers['content-type']` output
text/html; charset=UTF-8
halcyon plankBOT
trim fiber
I am sorry I am gonna ask more questions on that lol
Go ahead!
trim fiber
I don't understand, you have r = requests.get(...) right?
unique pecan
Yeah
trim fiber
So try to execute the following code
import json
data = r.text
data = json.loads(data)
print(x)
``` return
file_engines
domain_engines
if that means anythingIts the keys to the response dict
trim fiber
```for x in r.json():
print(x)
``` return
file_engines
domain_engin...
Okay, so when your data is
{
"key": [
"value 1",
"value 2"
]
}
You can process it by
for key, values in r.json().items():
print("key", key)
print("values", values)
Awesome thanks!
unique pecan
import requests
url = "https://www.scanlabs.net/api/process-domains?username=&password="
payload='selection%5B0%5D=44&selection%5B1%5D=43&entry%5B0%5D=google.com&entry%5B1%5D=google.in'
headers = {
'Content-Type': 'application/x-www-form-urlencoded'
}
response = requests.request("POST", url, headers=headers, data=payload)
print(response.text)
Those two are same thing ^
how do I POST a URL?
I dont follow
old rover
```py
import requests
url = "https://www.scanlabs.net/api/process-domains?user...
hey this would be more suited for #web-development
wet urchin
Hey guys
Can you recommend resources to learn Data Structures? It could be webpages, youtube channels or books. Thanks in advance
old rover
Can you recommend resources to learn Data Structures? It could be webpages, yout...
this guy has good youtube videos on algorithms
wet urchin
Thanks, do you have another for data structures?
old rover
I didn't have one single source for data structures
I had to hop around a bit
I also did the Udacity Python DS/algos course but it was in Python 2.7
it wasn't that bad tho
I also used Grokking Algorithms
also before you go into ds/algos you should know OOP well
at least the basics of OOP
if you need help w OOP you can check out Corey Schafer's youtube series on OOP
vital rune
im trying to search for patterns in a trie is anyone familiar with tries?
agile sundial
what do you mean by patterns?
boreal stratus
what's a way to test whether a word is in alphabetical order?
gritty marsh
what's a way to test whether a word is in alphabetical order?
Map each letter to its ascii code and perform numeric comparisons
vocal gorge
Map each letter to its ascii code and perform numeric comparisons
no need, < works on strings too.
gritty marsh
Good to know
fiery cosmos
Similar to leetcode 437.
This algorithm is to find out the total number of paths in a binary tree whose sum is S. The path doesn't have to be only root to leaf node.
1 def count_paths_recursive(currentNode, S, currentPath):
2 if currentNode is None:
3 return 0
4
5 # add the current node to the path
6 currentPath.append(currentNode.val)
7 pathCount, pathSum = 0, 0
8 # find the sums of all sub-paths in the current path list
9 for i in range(len(currentPath)-1, -1, -1):
10 pathSum += currentPath[i]
11 # if the sum of any sub-path is equal to 'S' we increment our path count.
12 if pathSum == S:
13 pathCount += 1
14
15
16 pathCount += count_paths_recursive(currentNode.left, S, currentPath)
17
18 pathCount += count_paths_recursive(currentNode.right, S, currentPath)
19
20 # remove the current node from the path to backtrack
21 # we need to remove the current node while we are going up the recursive call stack
22 del currentPath[-1]
23
24 return pathCount
can anybody help me why pathCount returns the total path since it keeps changing to 0 at line 7. I have been stuck in this since yesterday and couldn't find an explanation.
atomic folio
Can anyone help me with a variable size sliding window problem
cloud plover
I need to show something on a share screen it is pygame and algorithm related can someone help me with that?
I will explain but it's hard to say what i mean in text without sharing my screen
hybrid zodiac
HEYYYYY
can i get help with very simple python
its on a website called cmu academy
old rover
!rule 5
halcyon plankBOT
manic swift
could someone help me with returning the empty list? not sure why im getting (0,0)
steep lily
Can someone help me understand Expectimax and the implementation of it as compared to minimax with pruning?
I can hop in a vc
grizzled schooner
could someone help me with returning the empty list? not sure why im getting (0,...
you do your check for an empty list within the for loop, but the for loop will never even happen if the list is empty
manic swift
you do your check for an empty list within the for loop, but the for loop will n...
thanks a lot it worked
grizzled schooner
np!
manic swift
np!
im stuck with another problem
error:
AttributeError: 'list' object has no attribute 'replace```
im trying to remove vowels from the string
i dont know if my attempt at list comprehension is even right, but i thought it would allow me to use the replace method?
spring jasper
2 lines up, string is made a list
manic swift
so here:
new_string = [string.replace(letter, "") for letter in string if letter == vowel]
is this the issue?
spring jasper
You could also use str.translate
string.translate({ord(vowel):"" for vowel in "aeoiuAEOIU"})
sorry i dont undestand how to fix this problem
agile sundial
no, the issue is that splitting a string gives you a list
grizzled schooner
~~no, the issue is that splitting a string gives you a list~~
i donβt see them splitting a string anywhere
agile sundial
mmm, the issue is in my reading skills
grizzled schooner
indeed
agile sundial
the actual issue in the code is because you aren't using your loop variable
spring jasper
also, str.replace replaces all occurrences of the letter, you dont need to loop over the letters in the string
manic swift
what if i try this:
for vowel in vowels:
if vowel in string_:
string_.replace(vowel, "")```You also dont need to check beforehand, str.replace wont error if the letter isnt in the string
Also you gotta catch the return
manic swift
so i dont need the if vowel in string_: ?
i tried printing string_ but the string stays the same it doesn't change at all
spring jasper
An example #bot-commands message
manic swift
thanks it worked, looks like i was just complicating it for no reason π₯
manic swift
An example https://discord.com/channels/267624335836053506/267659945086812160/83...
could u explain one last thing, in string_.translate(None, "aeiouAEIOU") , why do you put None ?
spring jasper
Uh, thats not what the code says
manic swift
oh yeah, after submitting it shows other peoples code and i got that from there ^, since u were talking about translate() before i just picked that code.
spring jasper
If you want to use str.translate you have to give it a mapping/dict of the unicode code for a particular letter and what you want it to replace
#bot-commands message
Kinda sad how i cant eval in here
Or can I
!e
halcyon plankBOT
You are not allowed to use that command here. Please use the #bot-commands channel instead.
spring jasper
Dang, would be useful
golden moat
>>> 2023049.2369
2023049.2369
>>> 2023049.236 + 9 * math.pow(10, -4)
2023049.2369000001
any way to avoid that pow returns 0.0009000000000000001 and instead returns 0.009?
feels like I'm missing something very basic about floats
spring jasper
dapper sapphire
oh wow lol that's an actual link.
halcyon plankBOT
native oar
hi, i have implemented breadth first search to work out the shortest path in a graph by number of edges, and dijkstra for shortest path based on weight. Now how would i test empirically their order of growth, since bfs is O(V + E) and dijkstra should be O(ElogV) if i am not mistaken using timeit? I created test graphs, where the number of edges is constant but the number of vertices vary and tried to measure time with increasing V. In both cases they give me a form of linear relationship when i plot the data using pyplot
agile sundial
testing empirically will always be worse than proving theoretically
how large are your test graphs?
you're probably not going to notice a difference between bfs and dijkstra's until like 20k nodes probably
native oar
Not very large, have 5 edges, and vertices range from 100 to 20000, not sure if thats how you are supposed to empirically determine the order of growth, like the method i am using. Is it maybe because i have such a low number of edges compared to vertices? Is the method wrong? What would i need to measure in that case to test it emprically?
icy spruce
@lean dome I thought big theta only applies if big O and omega are the same, otherwise it wouldn't exist? Correct me if I'm wrong
lean dome
Big o is an upper bound, but not necessarily a tight upper bound.
Big omega is a lower bound, but not necessarily a tight lower bound.
We usually drop the constants off of those, and talk only about tight bounds.
Every algorithm that's O(1) is also O(n) and O(n!), according to a mathematician or a CS professor.
According to anyone you talk to in industry, they'll only call it the tightest one of those, O(1)
icy spruce
So based on definition there are algorithms where big theta doesn't exist if tight O and tight omega are different?
From CtCI book:
(big theta): In academia, big theta means both O and omega. That is, an algorithm is theta(N) if it is both O(N) and
omega( N). Theta gives a tight bound on runtime.
half lotus
Yes, it doesn't always exist is how I learned it.
The upper and lower bound functions don't have to be exactly identical; they can differ by some constant value.
lean dome
By some constant factor, that is. If the number of operations performed for sufficiently large inputs is always greater than 4 * 2^n and less than 10 * 2^n, then it's big theta of 2^n
lean dome
The formal definition of big theta, per Wikipedia, is that a function f(n) is ΞΈ(g(n)) if there exist some positive k1 and k2 such that, for all n greater than some n0, k1 * g(n) β€ f(n) β€ k2 * g(n)
acoustic shadow
How do you use configparser?
glossy crypt
does anyone know recurrence systems (induction proofing)?
vocal gorge
what's your question?
kindred coyote
hey
so im doing some practice questions and i found this problem
can anyone help me?
ping me if u can
old rover
what's the question
old rover
is this a school assignment
kindred coyote
no
it's a sample problem
i saw online
idk how to really do it. im tryin to improve
here's sample input:
so far i worked out how to make the graph
i was thinking to perform a DFS on it, But that didnt work out
@old rover any progress?
kindred coyote
the problem is above
spring jasper
no i mean a link to it
kindred coyote
do u want the pdf to it?
spring jasper
oh its a pdf? i thought it was an online thingy
spring jasper
nah its fine, we got the screenshots
kindred coyote
ok
this is my solution: it doesnt work for all test cases: ```py
import sys
from collections import defaultdict
raw_input = sys.stdin.readline
n = int(raw_input())
graph = defaultdict(list)
for i in range(n-1):
pair = raw_input().split()
graph[pair[0]].append(pair[1])
graph[pair[1]].append(pair[0])
hc = raw_input().split()
def dfs(graph, node, path, visited=None, i=0):
if visited is None:
visited = set()
if node not in visited :
path.append(node)
visited.add(node)
for neighbour in graph[node]:
dfs(graph, neighbour, path, visited)
return path
op = dfs(graph, hc[0], [])[dfs(graph, hc[0], []).index(hc[0]):dfs(graph, hc[0], []).index(hc[1])
]
sp = dfs(graph, hc[0], [])[dfs(graph, hc[0], []).index(hc[1]):]
if len(op) % 2 == 0 or len(sp) % 2 == 0 and (len(op) != 0 and len(sp) != 0):
print("Let's play >:)")
else:
print("NOU >:(")
wdym test cases, so is this online or not lmao
stoic citrus
guys how do i instantiate a numpy array of the form [[9.71158805]] which my program tells me is of shape 1,1
and append this value to the array
dapper sapphire
How do companies implement search query detector, so if you someone looked for:
BILY1234-789
And if it is not in the database, but BILY-1234-789 is then it would show that
vocal gorge
How do companies implement search query detector, so if you someone looked for:
...
So, fuzzy matching?
dapper sapphire
Oh it's called fuzzy matching
vocal gorge
yup, check out !pypi fuzzywuzzy for a working library, even
dapper sapphire
Is that what elastic search does? But elastic search is an engine
dapper sapphire
yup, check out `!pypi fuzzywuzzy` for a working library, even
Thanks!! Really helpful.
frank osprey
Can someone plsss help me with this?
"Describe how to implement a dictionary that has an average lookUp, insert, and delete time of O(1), but uses an array of no more than 2 times the maximum number of elements that will be stored in the dictionary"
I read that I could implement a hashtable but I am not sure
vocal gorge
dictionary that has an average lookUp, insert, and delete time of O(1),
that's pretty much only achieved by a hashmap, I believe
onyx root
hashmap is a Java term
vocal gorge
Not only, I think it's how they are called in C++, and from there in a lot of languages
frank osprey
I am supposed to do it in java anyway, so thank you!
serene oracle
C++ calls it unordered_map but it's all the same concept
unique pecan
Hi
faint sentinel
hello , i am working on min notes needed problem , but my code is in c , can i paste it here somehow
steep lily
Kinda depends. If you need to be indexing and need order in your data structure list is the way to go. Dict are unordered and accessible with keys(which most of the time I prefer)you can easily use a key named anything if you don't wanna deal with integers.
With lists, if it's a large data with lot of elements trying to remember the position of each element with the index values might not be super ideal. But if it's something small and simple lists are the way to go to loop over.
agile sundial
trying to remember the position of each element
since when do you need to do this? the idea of a list is that each element in it is roughly the same. there's no semantic difference between each element
steep lily
umm. in a dict there isn't any first element or second or nth element. Everything is just directly accessed by the key-value. in a list each element has a certain position, like 1st second, third ....nth
agile sundial
dictionaries are used for when you need a mapping between two things, for example a teacher's name to a list of their students
in python 3.7+, dictionaries maintain their insertion order
steep lily
That's also why dicts are faster than lists in large data since traversing through a dictionary doesn't take long. There is more space taken for a certain element to be accessed via it's position in lists.
agile sundial
traversing a dictionary is going to be the same or worse speed than a list
dictionaries are best at finding the specific value of a specific key
steep lily
traversing a dictionary is going to be the same or worse speed than a list
wait fr? I swear we did a comparison in a class and the dict was like 2 seconds or something faster.
steep lily
dictionaries are best at finding the specific value of a specific key
yes this.
Lookup vs iteration
dusk anchor
How do i make a algorithm that gets the letters of a given input, and then if an other input is equal to the letters stored, tell you the matches.
Example:
Word Input:
'Hello'
Variable called chars:
[('H', 'He','Hel','Hell')]
Search Input:
He
Output:
Did you meant Hello?
This is called prefix search. It's usually efficiently implemented with a trie.
dusk anchor
So how am i able to do it
vocal gorge
Like, yourself, or do you want a trie library?
dusk anchor
lib would be fine, but i would also like to get a basic knowledge on how to do it my self.
vocal gorge
See https://en.wikipedia.org/wiki/Trie, basically
vocal gorge
Tries are relatively simple in principle - it's a tree of characters you descend to find (or not find) the words
dusk anchor
like i want that for a discord bot.
dusk anchor
Tries are relatively simple in principle - it's a tree of characters you descend...
yea
halcyon plankBOT
vocal gorge
it has very horrible docs, sadly, but it doesn't have many functions, either, so you can just read the stubs
dusk anchor
Okay
vocal gorge
it really is performant, though
dusk anchor
For any external problems can i contact you?
vocal gorge
just ask somewhere
dusk anchor
Thanks.
vocal gorge
and if you want fuzzy search instead, like Maroloccio suggested, fuzzywuzzy
dusk anchor
WHAT THE FUCK ARE THESE DOCS LMAO
vocal gorge
that's more common for search suggestions
dusk anchor
yea
need to print that lmao:
Using
string = 'Hello'
char_list = []
for char in string:
char_list.append(char)
I am getting
['H','e','l','l','o']
How am i able to get
['H','He','Hel','Hell','Hello']
@vocal gorge
vocal gorge
well, append a slice of the string each time, for example
dusk anchor
hm like
for char in string:
char_list.append(char + char)
I am getting a double input at the list
['HH','ee']
etc
trim fiber
You can use range(len(string)) to generate list of possible indexes and then use string[:x] to have substring of string of length x
!e
string = "Hello"
x = 3
substring = string[:x]
print(f"x = {x}, substring = {substring}")
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
x = 3, substring = Hel
Rather
for i in range(len(string)):
# todo create x from i
pass
string = 'Hello'
x = 0
for char in string:
x+=1
print(substring[:x])
i think i can sue that
use*
This gives as output
H
He
Hel
Hell
Hello
Now that i am using this:
string = 'Hello'
char_list = []
x = 0
for char in string:
x+=1
char_list.append(string[:x])
print(char_list)
I am getting the expected output
['H', 'He', 'Hel', 'Hell', 'Hello']
Now how do i make it NOT to append the actual string.
To not append the Hello so the output will be:
['H', 'He', 'Hel', 'Hell']
@trim fiber
I am going to try:
if x == len(string)
is there a reason you're iterating over the string, rather than a range? you never use the value char
dusk anchor
It works fine with me
just want to not append the actual string at the table
yeah i made it
string = 'Hello'
string_length = len(string)
char_list = []
x = 0
for char in string:
x+=1
if x != len(string):
char_list.append(string[:x])
print(char_list)
Output
['H', 'He', 'Hel', 'Hell']
yes, it works fine, but it's not the best because you have unused variables. you could even avoid having to do the if x != len(string) check if you used range()
grizzled schooner
well, which range do you want x to be iterating over
think about the values of x in your current for loop (starts at 0, then you add 1, etc)
dusk anchor
yeah
I think i made it my self!!!
string = str(input('Enter a key: '))
string_length = len(string)
char_list = []
x = -1
for char in string:
x+=1
if x != len(string):
char_list.append(string[:x])
print(char_list)
search_input = str(input('Enter a keyword: '))
if search_input in char_list:
print(f'Do you mean {string}?')
@grizzled schooner @trim fiber
Tell me your opinion
trim fiber
I think i made it my self!!!
```py
string = str(input('Enter a key: '))
string_l...
First of all when you set x = -1 and then update it x += 1 you have x = 0 so you put empty string into char_list
!e
for string in ["abc", "test", "any string"]:
print(f"{string}[:0] = {repr(string[:0])}")
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
001 | abc[:0] = ''
002 | test[:0] = ''
003 | any string[:0] = ''
Secondly you don't need to create list of substrings because you can use .startswith method
dusk anchor
First of all when you set `x = -1` and then update it `x += 1` you have `x = 0` ...
i did x = -1 to get the first character
trim fiber
i did x = -1 to get the first character
Then set to 0 because as you can see you put empty string
dusk anchor
oki
trim fiber
@dusk anchor check the following example and think how can you calculate x from i
for i in range(len(string)):
print(string[:i])
okay
string = 'hello'
string_length = len(string)
char_list = []
x = -1
for char in string:
x+=1
if x != len(string):
char_list.append(string[:x])
search_input = str(input('Enter a keyword: '))
if search_input in char_list:
print(f'Do you mean {string}?')
And works, but how am i able to do that for multiple words. Like:
string = ['hello', 'test word']
@grizzled schooner
spring jasper
What is that even for
fiery plank
I wanna get started with algos, anyone know some really easy ones?
agile sundial
What is that even for
fuzzy matching?
spring jasper
Maybe im just bad at reading other people's code but it looks like it just produces all the prefixes of the string
main flower
I wanna get started with algos, anyone know some really easy ones?
It depends on what you are looking for. You could e.g. google cp-algorithms and you'll find there many nice algos and data structures. imo the best way to learn algos and DS is competitive programming. This way you can fully understand the functionality of certain algos/DS
But it depends on what you need
fiery plank
Hmm gotcha how would I even get into competitive programming
spring jasper
You start by learning some basic data structures and algorithms
Then try solving puzzles
Practice makes perfect
fiery plank
ok thanks
main flower
There are many online judges out there e.g. Spoj, Codeforces, uva, etc.
Then, just start solving problems
Cuz knowing algos/DS without knowing how to use them is pretty much pointless
onyx agate
Anyone have familiarity with Merkle Hash Trees? I'm trying to understand how they grow.
Specifically when they aren't "perfect binary trees"
prime kettle
Hey, how's going ? I hope everything is okay
aliaga = [
{"hour": "10:00", "amount": 100},
{"hour": "12:00", "amount": 120},
{"hour": "08:00", "amount": 150},
{"hour": "07:00", "amount": 150}
]
I have a array and I want to sort by hour's value.
Can someone give a suggestion ?
Thanx
grizzled schooner
Hey, how's going ? I hope everything is okay
```py
aliaga = [
{"hour": "10:0...
.sort() has a key kwarg. do you know about lambdas?
prime kettle
`.sort()` has a `key` kwarg. do you know about lambdas?
from datetime import datetime
aliaga.sort(key=lambda x: datetime.strptime(x["hour"], "%H:00"))
This returned None
grizzled schooner
yes, sort() doesn't return anything, it sorts the list inplace
prime kettle
Ohhh, you're right
grizzled schooner
sorted() returns a new sorted list, if that's what you're looking for
prime kettle
Thank you, it's fixed
grizzled schooner
happy to help~
stable lynx
Pls see #π€‘help-banana my algorithm is going slower than a linear search and not sure why
austere jackal
Hi,
Is this the right place to ask a question regarding data structure?
I'll go ahead and post my question just in case. If I'm in the wrong spot just let me know and I'll move it.
This isn't my actual data, but it gets the idea across. I'm trying to change data similar to DATASET 1 to look more like DATASET 2 using pandas.
I was wondering if there is a good way to make this adjustment using pandas.
My actual data has about 25 * 12 * 2 variables (25 variables for 12 categories with 2 sections each).
urban tulip
what are differences between recursion and divide-and-conquer?
agile sundial
divide and conquer is a way to use recursion
urban tulip
recursion is part of divide-and-conquer?
brave oak
recursion is part of divide-and-conquer?
no
fiery cosmos
more like the other way around. Divide and conquer is an idea you might use when writing a recursive function
divide the problem up into two subproblems, call them left and right, recurse on left, recurse on right
fiery cosmos
same type
Like with merge sort you split the list up into a left and a right and you recursively sort each
urban tulip
is this also called divide-and-conquer: original problem --> subproblems of different types --> subproblems of same type?
fiery cosmos
Have a problem in mind that does that? 
Though I'd say sure. If it divides the problem up into more manageable bits then it could be described as "divide and conquer".
It's just a descriptor , not a super technical term
agile sundial
divide and conquer is when you make subproblems, recursively solve them, then put the results together to solve the whole thing
and yeah, it's not a super well defined term like he said
urban tulip
i think i got it more clear, thanks discretegames and public static void
dusk anchor
string = 'hello'
string_length = len(string)
char_list = []
x = 0
for char in string:
x+=1
if x != len(string):
char_list.append(string[:x])
print(char_list)
search_input = str(input('Enter a keyword: '))
if search_input in char_list:
print(f'Do you mean {string}?')
How am i able to make this work for multiple strings
string = ['word1', 'word2', 'word3']
@trim fiber
trim fiber
```py
string = 'hello'
string_length = len(string)
char_list = []
x = 0
for ch...
Have you tried .startswith method?
if string.startswith(search_input):
pass
You don't need char_list anymore when you are using .startswith
dusk anchor
hm im kinda confused
trim fiber
!d str.startswith
halcyon plankBOT
dusk anchor
what this does is checking if the string , starts with the input
trim fiber
This method do exactly what your char_list do
dusk anchor
char_list just contains the possibilities:
string = 'hello'
char_list = ['h','he','hel','hell']
but starts with just checks if the string starts with the input
crystal goblet
hi
dusk anchor
hello
trim fiber
char_list just contains the possibilities:
```py
string = 'hello'
char_list = ['...
But all variants inside char_list are just first part of the string
I mean - you are creating substring by cutting few first letters
dusk anchor
yeah. I mean i need that for my discord bot.
there is a table that stores the commands
commands = ['command1', 'command2']
And if the user types !help commnd, for example it will say: Command not found, Did you meant {the command that matches}
trim fiber
So when you have
commands = ["help", "hello"]
and you put he it should return only help or both commands?
dusk anchor
it will return both.
trim fiber
Do you know [x for x in y] notation?
dusk anchor
But if the case was like:
commands = ['report', 'suggest']
And you put repo, it will return report, because this matches better.
halcyon plankBOT
dusk anchor
okay so
commands = ['command1', 'command2', 'command3']
hints = []
?
like
what is the alg to get the "hints"
trim fiber
So how you create single hint when you have commands and search_input?
dusk anchor
alright..
trim fiber
commands = ['command1', 'command2', 'command3']
search_input = str(input("put your command"))
hint = ...
yeah
trim fiber
You already know that you can use .startswith
dusk anchor
yeah but
startswith will give the whole command
if search_input.startswith(commands)
Nope, in different way
for command in commands:
if command.startswith(search_input):
...
!d str.startswith
halcyon plankBOT
dusk anchor
okay so
trim fiber
It checks whether string starts with given prefix
dusk anchor
for a value in commands table, if the value startswith what the user inputted.
yeah its good
trim fiber
!e
string = "Hello world!"
print("starts with 'Hello'?", string.startswith("Hello"))
print("starts with 'hello'?", string.startswith("hello"))
print("starts with 'world'?", string.startswith("world"))
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
001 | starts with 'Hello'? True
002 | starts with 'hello'? False
003 | starts with 'world'? False
As you can see it's also case sensitive
dusk anchor
yea
commands = ['reprot', 'call']
search_input = str(input('Enter a command: '))
for command in commands:
if command.startswith(search_input):
print(f'Did you mean {command}')
Got thqat
Got this
works fine
i meant report not reprot xD
trim fiber
Okay, can you make function that for given commands and search_input gives list of possible commands?
trim fiber
Like
commands = ["report", "call", "help"]
search_input = "re"
hints = your_nice_function(commands, search_input)
print(hints) # returns ["report"]
yeah
let me try
i made one
let me try
made it
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
print(possible_commands)
check(commands_list, search_input)
@trim fiber
['report']
Your function doesn't return anything
dusk anchor
it does
trim fiber
!t return
halcyon plankBOT
dusk anchor
!eval
commands_list = ['report', 'call', 'hello']
search_input = 're'
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
print(possible_commands)
check(commands_list, search_input)
@dusk anchor :white_check_mark: Your eval job has completed with return code 0.
['report']
see?
dusk anchor
!eval ```py
commands_list = ['report', 'call', 'hello']
search_input = 're'
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
print(possible_commands)
return possible_commands
check(commands_list, search_input)
!eval
commands_list = ['report', 'call', 'hello']
search_input = 're'
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
print(possible_commands)
return possible_commands
check(commands_list, search_input)
halcyon plankBOT
@dusk anchor :white_check_mark: Your eval job has completed with return code 0.
['report']
You have typo in indentation
commands_list = ['report', 'call', 'hello']
search_input = 're'
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
# print(possible_commands) <- this line is no needed
return possible_commands # <- this line had wrong indentation
check(commands_list, search_input)
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
return possible_commands
hints = check(commands_list, search_input)
print(hints)
now it works
dusk anchor
You have typo in indentation
```py
commands_list = ['report', 'call', 'hello']
s...
in my ide it is fine
dusk anchor
```py
def check(commands, search_input):
possible_commands = []
for comm...
see that
@trim fiber
trim fiber
now it works
For commands = ["hello", "help"] it returns only one possible command, not every possible
dusk anchor
hm
trim fiber
!e
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
return possible_commands
hints = check(["hello", "help"], "he")
print(hints)
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
['hello']
!e
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
return possible_commands
hints = check(["hello", "help"], "he")
print(hints)
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
['hello', 'help']
Like I said, indentation of return is wrong
dusk anchor
okay
so i should use
def check(commands, search_input):
possible_commands = []
for command in commands:
if command.startswith(search_input):
possible_commands.append(command)
return possible_commands
hints = check(["hello", "help"], "he")
print(hints)
If you want to return all possible variants - yes
halcyon plankBOT
dusk anchor
afk
trim fiber
Like
nums = [0, 1, 2, 3, 4, 5, 6, 7]
evens = []
for num in nums:
if num % 2 == 0:
evens.append(num)
Shorter form
nums = [0, 1, 2, 3, 4, 5, 6, 7]
evens = [num for num in nums if num % 2 == 0]
okay
so
commands = ['command1', 'command2']
hints = [command for command in commands if command.startswith(search_input')
trim fiber
commands = ['command1', 'command2']
hints = [command for command in commands if ...
That's right
dusk anchor
lmao it works
search_input = 'he'
commands = ['hello', 'help']
hints = [command for command in commands if command.startswith(search_input)]
print(hints)
calm wraith
Can someone explain binary search to me? I've been told to use it in my program but I don't understand it entirely, thanks.
main flower
So, let's say you have a sorted (non-decreasing) array of length n. Then, you can easilly see that If i and j are indexes of elements then if i > j then obviously array[i] >= array[j].
Now, let's get to binary search. You calculate the index of middle element and see If the number you are looking for is bigger than or equal to this middle element or not. If it is, It's obvious that you won't need to look through elements smaller than middle element. So now you calculate the index of middle element in that smaller subarray and so on untill you find the number you were looking for.
a = [1, 3, 6, 8, 10, 18]
def bin_search(a, num, l, r):
mid = (l+r)/2
if a[mid] == num:
return mid;
if a[mid] > num:
bin_search(a, num, mid, r)
else:
bin_search(a, num, l, mid)
bin_search(a, 3, 0, len(a)-1)
Unless I did some pretty stupid mistakes in this code xd it should work
You can also google How does binary search work and I'm sure you will find many good visualisations of it
frosty igloo
What would be the fastest way to apply numpy.linalg.norm() to each value of an 2d array with a given value x?
dusk anchor
command_input = ''
commands = ['report', 'call']
hints = [command for command in commands if command.startswith(command_input)]
print('Did you mean,',hints)
When i am using that ^^
I am getting the following output:
Did you mean, ['report', 'call']
How can i make my output to appear as:
Did you mean, report - call ?
@trim fiber
trim fiber
```py
command_input = ''
commands = ['report', 'call']
hints = [command for com...
Have you seen str.join?
!d str.join
halcyon plankBOT
trim fiber
!e
print("x".join(["a", "b", "c"]))
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
axbxc
!eval
command_input = ''
commands = ['report', 'call']
hints = [command for command in commands if command.startswith(command_input)]
print('Did you mean,',hints.join())
hm
trim fiber
Itβs separator.join(hints)
dusk anchor
so
like
print('Did you mean,',' - '.join(hints))
print('Did you mean,',' - '.join(hints))
!eval
command_input = ''
commands = ['report', 'call']
hints = [command for command in commands if command.startswith(command_input)]
print('Did you mean,',' - '.join(hints))
@dusk anchor :white_check_mark: Your eval job has completed with return code 0.
Did you mean, report - call
yeah
alright
hm okay now
i need some help on applying that
to my bot
let me try it first
fiery cosmos
is this code good?
from pycoingecko import CoinGeckoAPI
import webbrowser
# initializing the api
cg = CoinGeckoAPI()
# created a function to check for the price of STX
def check_price():
# give the api an id and a fiat currency
price_dict = cg.get_price(ids='blockstack', vs_currencies='usd')
# check price
price = price_dict['blockstack']['usd']
# if it's 3$ open a web browser window
if price == 3:
webbrowser.open("https://www.coingecko.com/en/coins/stacks")
else:
# else redo
check_price()
check_price()
What do you mean?
spring jasper
That code looks like it would give indentationerror
Also its not really ds/algo related
fiery cosmos
i think my code is formated
trim fiber
is this code good?
```python
from pycoingecko import CoinGeckoAPI
import webbrow...
Few things:
Pass cg as an argument
You have recursive calls
fiery cosmos
Also its not really ds/algo related
i would like to know if this passes the clean code test
i think it looks bad
trim fiber
i think my code is formated
My fault then, I donβt see colors on mobile Curious
dusk anchor
is this code good?
```python
from pycoingecko import CoinGeckoAPI
import webbrow...
Please keep following the topic of the channel
spring jasper
A help channel maybe
fiery cosmos
A help channel maybe
thanks
dusk anchor
Try posting it in cocurrencies
Or at a help channel
@trim fiber check #discord-bots
trim fiber
<@!690641855045697653> check <#343944376055103488>
Easy
dusk anchor
yeah i just dont remember
trim fiber
is this code good?
```python
from pycoingecko import CoinGeckoAPI
import webbrow...
What is the real problem?
dusk anchor
fiery cosmos
What is the real problem?
it runs for a few seconds and it blows up
for no reason
or i'm too stupid to know the issue
trim fiber
I mean that I don't understand what is going on when you run the program and what do you expect
spring jasper
Because you have recursive calls as @trim fiber said
Its probably destroying your browser everytime you call it
fiery cosmos
I mean that I don't understand what is going on when you run the program and wha...
i expect this, it asks the api for the current price of STX if the price is 3 dollars i want it to open a browser window, if it's not 3$ i want it to repeat itself
i fixed it
from pycoingecko import CoinGeckoAPI
import webbrowser
import time
# init api
cg = CoinGeckoAPI()
# a function to check for the price
def check_price():
price_dict = cg.get_price(ids='blockstack', vs_currencies='usd')
# getting the price in form of an int
price = price_dict['blockstack']['usd']
# keep doing this forever
while True:
# if the price is 3 usd
if price == 3:
# open a web browser with the provided link and break the loop
webbrowser.open("https://www.coingecko.com/en/coins/stacks")
break
# else show me where we're and wait a minute to redo the thing
else:
print("price is at", price)
print("--------------------")
time.sleep(60)
check_price()
basically
explained
spring jasper
That will only check the price once and keep the loop going forever if the first time isnt 3
Without checking again
Why not
while not price == 3:
print("price is at", price)
print("--------")
time.sleep(60)
price_dict = cg.get_price(...)
price = price_dict["blockstack"]["usd"]
webbrowser.open(...)
big bren
thanks man
i'm still a noob
tiny river
Is it possible to have a pointer to the next index in the numpy array? I wonder if I can do something faster with ctypes than a regular index +=1;array[index] = something
vocal gorge
Perhaps if you use a raw pointer to access the array. They are contiguous in memory, so you'd increment the pointer by the width of the dtype to get the next element.
No idea how to access numpy arrays using raw pointers, though.
lean dome
Cython can do that.
kindred coyote
given the tree 8 4 \ / 6 3 1 2 / \ 7 5. Starting at node 2, how would you traverse the path(s) 2 -> 4 -> 5 and 2 -> 1 -> 3 -> 6 -> 8 -> 7?
keen hearth
given the tree ```
8 4
\ /
6 3 1 2
/ \
7 5
```....
Hey! I'm not sure I understand the question. Are you simply looking to do a depth-first search of the tree, starting at 2?
dapper sapphire
So is there a way of knowing end of a string in python? so something similar to in C where we detect the null terminator:
while(string[i] != '\0')
{
}
Uhh, are you asking about some way of directly accessing the underlying representation of the string with ctypes?
Because if you mean just normal Python strings, they act like lists of their characters. You'll get an IndexError for trying to access after the end.
dapper sapphire
actually I can just len(string)
vocal gorge
(so basically, Python strings are arrays of characters with known lengths, so there's no null terminator necessary)
dapper sapphire
Because if you mean just normal Python strings, they act like lists of their cha...
Yeah, that's what happend
That was one of the strangest things I noticed about python, string are immutable
so if we have:
string = "abcdef"
we cant do:
string[0] = 'A'
But I think python isnt the only language where strings are immutable.
vocal gorge
I believe one of the reasons is that this way you can safely use strings as keys of dicts
I often just do s = list(s) and then s = "".join(s) when I have to edit a string by char.
agile sundial
java has immutable strings
dapper sapphire
I believe one of the reasons is that this way you can safely use strings as keys...
Actually that's a pretty good and valid point.
vocal gorge
if strings were mutable by default, there'd have to also be a frozenstr class (like there's frozenset) and you'd have to cast any string you want to use as a key to that class
agile sundial
that wouldn't actually be that bad though
vocal gorge
also, there would be many questions like
s = "hello"
res = []
cur = ""
for char in s:
cur += char
res.append(cur)
"Why am I getting the same string 5 times"?
as we are getting with lists currently π
vocal gorge
what language is that a joke about? Haskell?
agile sundial
there's frozenset for immutable so why not melted for mutable
vocal gorge
otn a meltedstr
agile sundial
π
vocal gorge
...moltenstr?
agile sundial
don't melted and molten mean the same thnig but just different connotations
molten does sound better though π€
vocal gorge
I've looked up how hashmaps work and for some reason the implementations I find use an array of buckets, with the hash being used to select an index in that array (which naturally has the disadvantage of requiring to preallocate all the buckets beforehand). Don't hashmaps usually use an approach like a prefix tree instead, where your hash decides a path along a binary tree to a bucket, which is a leaf?
agile sundial
i think most implementations will be very simplified for a school course. did you look at python's hashmap?
vocal gorge
hmm, guess I should finally watch https://www.youtube.com/watch?v=p33CVV29OG8 lol
agile sundial
ooh, that's a good one
vocal gorge
I think the problem is that a hash table is the name for the array implementation I found
lean dome
I've looked up how hashmaps work and for some reason the implementations I find ...
which naturally has the disadvantage of requiring to preallocate all the buckets beforehand
Why is that a disadvantage? Dynamic array elements are also preallocated, and that's the biggest advantage of a dynamic array - it's the thing that gives dynamic arrays amortized O(1) appends, as well as O(1) retrievals.
Don't hashmaps usually use an approach like a prefix tree instead, where your hash decides a path along a binary tree to a bucket, which is a leaf?
Not that I've ever heard of. You could use a trie to implement a mapping, but the end result wouldn't be a hashmap - it would have different average and worst case performance for most operations. Better worst case performance, worse average case performance.
vocal gorge
not sure about worse average case. Searching a tree to get to the bucket is O(depth), but since the values of a hash function are fixed (or at least bounded) size, the depth is limited, so it's O(1).
lean dome
hm - no, because you still have hash collisions to contend with. The worst case is still that every key you insert hashes to the same value, so the worst case is still O(n)
well, hash tables have a worst case of O(n), tries used to implement a hash map would have a worst case of O(n+m), where n is the number of elements in the hash table, and m is the length of the hash code.
vocal gorge
That's still O(n) if the length of the hash is bounded. Basically, the advantage I'm thinking of is that having the buckets be leaves of a search tree results in there automatically being a bucket for each (encountered) value of the hash function, rather then some predefined number that must be then grown.
lean dome
that would, in practice, be slower.
fiery cosmos
could someone point me to an article about implementing bloom filters? I don't want the code, I want to do it myself.
I prefer something close to pseudocode vs just a description
lean dome
you're right that it would have the same asymptotic complexity, I suppose - but at the cost of a fixed overhead of 64 extra indirections per lookup even in the best case, compared to the 1 indirection for looking up an element from a traditional, array-based hash table.
fiery cosmos
hm, seen <https://llimllib.github.io/bloomfilter-tutorial/>?
can a bloom filter be used for pattern matching on strings? like it'll tell me if a string "might be" in a set but never confidently that it is?
so if the failure rate is 2%, there is a 98% chance that the string is actually in the set?
lean dome
"pattern matching"? not really. Probabilistic lookups? yes.
fiery cosmos
i mean if i have 100 strings in the bloom filter, and then i have strings coming in one by one from a stream and i want to check each string from the stream as it comes in to check if it's in the set
i know there is a more optimal implementation called a "Cuckoo Filter"
I want to take more of a data structures approach to pattern matching. My other idea was to implement a concurrent aho-corasick algorithm but that's been done ad nauseum.
lol
also, we can't ignore the excellent space efficiency offered by a bloom filter
*assuming we accept a certain failure rate
lean dome
right - both bloom filters and cuckoo filters answer the question "is this element a member of the set", and for both the answer is either "definitely not" or "maybe"
fiery cosmos
could you verify that my understanding of the math is correct? "Maybe" means 98% chance assuming a 2% failure rate?
lean dome
there's no such thing a a "failure rate"... that's not really how it works at all. The more elements there are in the set, the more bits are set in the array, and the higher the rate of false positives going forward
that is, the failure rate changes as elements are inserted.
fiery cosmos
ah okay, i understand what you're saying. I see the formula now. how about I have multiple bloom filters that are pattern matched concurrently (asynchronously?) against incoming traffic? that would decrease the size of each bloom filter and increase the rate of success (probability that the maybe is actually a yes?)
i'm sorry if i'm not making sense but i can visualize it in my head
multiple bloom filters because the rate of false positives increases with n
by the way, thank you for listening and responding. i appreciate it.
lean dome
ah okay, i understand what you're saying. I see the formula now. how about I hav...
you could split the elements across multiple bloom filters to decrease the false positive rate, but it's easier to just keep one bloom filter and increase the size of its bit array, to reduce the odds of collisions within it
fiery cosmos
you could split the elements across multiple bloom filters to decrease the false...
oh, let me write that down!
thanks again, i'm going to go and study the data structure more thoroughly before i attempt an implementation.
native oar
hi, I would need some help solving a problem, in which i am given a set of vertices (places) and i need to return the shortest path (in terms of distance) connecting all vertices, now i've been told that it is similar to the travelling salesman problem, but the brute force approach would be inefficient, and in terms of heuristics to consider, I have been told to consider the euclidean tsp, but honestly i am not finding much about the euclidean version, I am not sure about what kinda implementation I would need to consider for the euclidean version. Any help would be appreciated regarding the euclidean and other heurestics that i can implement on a graph (i've implemented the edge weighted graph via adjacency lists)
agile sundial
it is similar to tsp, but you don't have to get back to the start point
fiery cosmos
I have a maybe weird question about sorting 
When you know the max size of the list and all the possible values of the numbers (or data) in it, couldn't you form a O(1) time sorting algorithm by making a huge lookup table for all possible permutations and their sorts that you calculate beforehand?
Requiring O(M^(N!)) space (I think?) where M is the range of values and N is max list size
no wait, just O(M^N) space
agile sundial
the time to create such a table would not be constant
and yeah, you could probably do that
fading sky
hello, I'm just getting started with DSA, where would you guys suggest to practice DSA problems? There are so many of them leetcode, hackerrank, hackerearth, etc. It's overwhelming.
fiery cosmos
hello, I'm just getting started with DSA, where would you guys suggest to practi...
I'd try LeetCode and GeeksForGeeks
fiery cosmos
and yeah, you could probably do that
I wonder if anyone ever actually does for very specific use cases 
azure gazelle
How do you fix this error when working with json in a for loop? RuntimeError: dictionary changed size during iteration
near sun
Is there anyone here,who currently working on python ?
Like an internship or job
Please dm me if u working on python
vocal gorge
When you know the max size of the list and all the possible values of the number...
The problem is that just looking up the sorted order for a given list in that table will be O(n), not O(1).
ivory yacht
You could do a hashmap. But memory requirements would quickly balloon to absurd levels.
vocal gorge
calculating the hash of a list of n elements is still O(n)
ivory yacht
Right, true.
dusk anchor
import string
letters = string.ascii_letters
commands = ['hello', 'help']
command_input = str(input('Enter a command: '))
def check(commands, command_input):
possible_commands = []
if command_input in commands:
print('Command found')
else:
for command in commands:
if command.startswith(command_input):
possible_commands.append(command)
print(possible_commands)
check(commands, command_input)
This works fine. I just want to check if the user inputted like:
'hellou'
#Or
'helloasasgawrsav'
So it will return again the best possible match.
@trim fiber
should i use:
x = len(command_input)
if x - len(command) == len(command):
if x.content == command:
?
soft totem
anyone know a good algorithm to find the prime number of a given number , without looping from 2 to n - 1 ??? like for finding the gcd we use euclid algo and for finding the prime number upto some n we use seive of eratos
anyone know a good algorithm ????
vocal gorge
...the Sieve of Eratosthenes, which you mentioned?
runic tinsel
anyone know a good algorithm to find the prime number of a given number , withou...
you mean factorize?
363 β 3, 11, 11
or what
soft totem
not factorize
runic tinsel
check primality?
soft totem
check primality?
meaning ???
mint jewel
do you mean generate all primes smaller than n?
runic tinsel
i don;t understand the question
soft totem
i wawnt to find whether the given number is prime or not , but i dont want to use that O ( n ) algo where we loop through 2, n-1 and check if its divisible by that num
glossy breach
You just have to loop from 2 to sqrt(n)
This will drastically make the program faster
runic tinsel
primality tests are 2 hard
glossy breach
Or more complicated, use miller-rabin primality test
soft totem
You just have to loop from 2 to sqrt(n)
this is to check whether the given num is prime or not right ?
glossy breach
Yes
soft totem
whats this algorithm called
glossy breach
I'm not sure if there's a name
soft totem
oh
correct me if i am wrong , so i have to loop from 2 , sqrt(n) and if it divisible by any of those num then its not a prime number
glossy breach
Yes
dusk anchor
```py
import string
letters = string.ascii_letters
commands = ['hello', 'help'...
so
any help
soft totem
@glossy breach ```
def prime(n):
isPrime = False
n = int(math.sqrt(n))
for i in range(2,n):
if n % i == 0 :
isPrime = True
break
return isPrime```
@soft totem can u help
soft totem
<@!393393798350372876> can u help
yeh what is it ?
dusk anchor
```py
import string
letters = string.ascii_letters
commands = ['hello', 'help'...
Here
runic tinsel
you can like manually skip over certain numbers
def isprime(n):
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
k = 5
m = 2
r = n ** .5
while k < r:
if n % k == 0:
return False
k, m = k+m, 6-m
return True
you reused the n variable
soft totem
you reused the `n` variable
oh yeh wawit , that was dumb of me
soft totem
you reused the `n` variable
def prime(n):
isPrime = True
for i in range(2,int(math.sqrt(n))):
if n % i == 0 :
isPrime = False
break
return isPrime```you need to add +1 to that range end
runic tinsel
you need to == 0
dusk anchor
@soft totem
soft totem
<@!393393798350372876>
wait gonna check
dusk anchor
ok
soft totem
ok
if user inputted like hellosssd then you want to check if the that inputted one has hello in it ???
and print that ?
dusk anchor
yeah
trim fiber
<@!690641855045697653>
Please, don't ping me every time when you need help
When I am able to help then I will respond
ashen glade
while choice =="yes":
import math
a = float(input("Enter Leading coefficient a"))
b = float(input("Enter the next coefficient b"))
c = float(input("Enter the constant c"))
if a==0:
print("Value of a should not be zero")
else:
D = b**2 - 4*a*c
if D>0:
R1=(-b+math.sqrt(D))/(2*a)
R2=(-b-math.sqrt(D))/(2*a)
print("Roots are Real and UnEqual")
print("Root 1 =",R1,"Root 2 =",R2)
elif D==0:
R1=(-b)/(2*a)
print("Roots are Real and Equal")
print("Root 1 =",R1,"Root 2 =",R1)
elif D<0:
R1=(-b+(math.sqrt(D))j/(2*a)
R2=(-b-(math.sqrt(D))j/(2*a)
print("Roots are Complex and Imaginary")
print("Root 1 =",R1,"Root 2 =",R2)
choice=input("Enter continue of not(yes/no)")
please tell me what should I do to get even output as complex numbers
soft totem
yeah
lol , had some work ```
def check(cmdIp, commands):
for cmd in commands:
if cmd in cmdIp:
print(cmd)
commands = ['hello','help']
cmdIp = input('Command : ')
check(cmdIp, commands)
check if this solves your question
fiery cosmos
calculating the hash of a list of n elements is still O(n)
Oh, good point. But still O(n) is better than nlogn
ashen glade
what are you getting right now?
I am getting error in the part after elif D<0:
ashen glade
(also, that belongs more in a help channel)
ohh I am actually new here so where is that help channel ?
fiery cosmos
steep lily
Can someone test my expectimax based game. I think I did a good job overall just wanted to get some feedback. I'll share the source code if you're actually down to review it too.
alpine pilot
I hope this is the right chat, how can I print in a readable way this dictionary?
{'streams': [{'index': 0, 'codec_name': 'h264', 'codec_long_name': 'H.264 / AVC / MPEG-4 AVC / MPEG-4 part 10', 'profile': 'Main', 'codec_type': 'video', 'codec_time_base': '1/48', 'codec_tag_string': 'avc1', 'width': 1080, 'height': 1080, 'coded_width': 1088, 'coded_height': 1088]}
half lotus
{'streams': [{'index': 0, 'codec_name': 'h264', 'codec_long_name': 'H.264 / AVC ...
No, not really the right channel. But to quickly answer, use pprint or json.dumps. If you have more questions about that, open a help channel #βο½how-to-get-help
alpine pilot
Sorry about that, thank you very much for the answer!
stable lynx
--need help----
Hey i have a quadtree search algorithm using random/normal data and for some reason its going slower than my linear serarch
And YES I have my start timer for both query searchers starting just before the actual query (so for example quadtree creation is NOT being added to the timer)
dusk anchor
lol , had some work ```
def check(cmdIp, commands):
for cmd in commands:
...
okay i will try.
wheat bear
hi, I would need some help solving a problem, in which i am given a set of verti...
Hello John, I would say that it depend on your goal. Are you looking for the best solution (then "careful brute force" like dynamic programming is a good solution) or a good solution (then meta-heuristics like simulated annealing are pretty simple to code and gives good results)? As far as I understand your problem, I believe that Euclidean or not does not change many things.
wheat bear
I'd try LeetCode and GeeksForGeeks
Hello, Can you tell me what DSA stands for?
half lotus
Hello, Can you tell me what DSA stands for?
Data structures and algorithms
dusk anchor
okay i will try.
yeah it works
dusk anchor
Hello, Can you tell me what DSA stands for?
Data Structures - Algorithms
commands = ['hello', 'help']
command_input = str(input('Command: '))
def check(commands, command_input):
possible_commands = []
if command_input in commands:
return 'Command found'
else:
for command in commands:
if command.startswith(command_input):
possible_commands.append(command)
return possible_commands
if command in command_input:
possible_commands.append(command)
return possible_commands
print(check(commands, command_input))
Any review on that?
fiery cosmos
ima need help you guys. I wanted to run this https://github.com/CorentinJ/Real-Time-Voice-Cloning on Archlinux and I installed everything except for the datasets because i wanted to clone datasets of the voice for GlaDos. But when running the demo_toolbox.py its non functional and i get this error in the terminal: QBasicTimer::start: QBasicTimer can only be used with threads started with QThread
this is what my interface looks like:
tranquil barn
Hi there! I needed help with this question https://www.codewars.com/kata/539a0e4d85e3425cb0000a88/train/python
I am not quite sure how to approach the problem
I can see the solution but I want to get a direction so I can try myself
soft totem
guys i have one question , if i got some solved answers and have some actual correct answers , is there any formula to get the percentage of my solved answers accuracy ?? like f1 score , but that one for ml
terse crypt
I can see the solution but I want to get a direction so I can try myself
the main issue here is that the result of add has to be a integer but also callable
now, how could one make a callable integer?
tranquil barn
the main issue here is that the result of add has to be a integer but also calla...
instance?
like instance of a class or any thing that can hold stuff i guess
structure
spring jasper
!e
class add:
def __init__(self, *args):
self.val = sum(args)
def __str__(self):
return str(self.val)
def __call__(self, other):
self.val += other
return self.val
def __add__(self, other):
return self.val + other
print(add(1,2,3))
print(add(1)(2))
print(add(1, 2)(3) + 6)
``` maybe?@spring jasper :white_check_mark: Your eval job has completed with return code 0.
001 | 6
002 | 3
003 | 12
I dont see how you'd implement this with partial functions
topaz pulsar
!e it's much easier to just subclass int ```py
class add(int):
def call(self, other):
return add(self + other)
assert add(1)(2)(3)(4) == 10
assert add(4) == 4```
halcyon plankBOT
@topaz pulsar :warning: Your eval job has completed with return code 0.
[No output]
lusty ivy
Whats the algorithm that pathlib.Path uses to iterate over a file system and mapping out nodes recursively?
like if i want to learn about doing that, what do i search
the binary tree ones seem limited to only left and right
topaz pulsar
it's quite similar to binary trees, file systems are a tree-like structure just not a binary one, you can apply the same algorithms, just you may find the children nodes are stored in a different way ```py
class Node:
... def init(self, value, children=()):
... self.children = list(children)
... self.value = value
...
def iterate(root: Node):
... stack = [root]
... while stack:
... node = stack.pop(0)
... stack.extend(node.children)
... yield node.value
...
...
root = Node(6, (
... Node(5),
... Node(4, (Node(3), Node(2))),
... Node(1),
... ))
print(*iterate(root))
6 5 4 1 3 2``` is BFS for example
@lusty ivy ^
if you wanted to do recursive searches you can implement something like DFS on a similar structure, I don't know of any particular resources to learn about what pathlib does exactly but in general, for non binary trees, you can just slightly modify your existing implementation to deal with however you are storing the variable number of child nodes
lusty ivy
Interesting.
Im confused by this a bit
>>> root = Node(6, (
... Node(5),
... Node(4, (Node(3), Node(2))),
... Node(1),
... ))
did you forget a close paren?
oh I get it
sort of haha. I gt everything up till that part
topaz pulsar
thats jut me creating a tree, the tree is structured 6 / | \ 5 4 1 / \ 3 2
jovial tartan
what's a good data structure for context-free grammars?
like if we have a dictionary of different types of words (i.e. <noun> can be 5 different words, <verb> can be 10 different), but there's also types such as <phrase-type-one> which consists of "<verb> the <noun>"
how would we store that?
i was thinking a tree but that would seem inefficient if we needed to have both a path to <verb> that doesn't lead from <phrase-type-one> and one that does lead from that. would a graph work instead? maybe a hashtable with separate chaining?
thanks!
fading sky
Hello, I'm just getting started with leetcode and this was the first problem presented to me related to arrays:
Given a binary array nums, return the maximum number of consecutive 1's in the array.
Example:
Input: nums = [1,1,0,1,1,1]
Output: 3
This was my solution at first :
def findMax(nums):
maxi = [0]
for i in range(len(nums)):
if nums[i]:
maxi[-1] += 1
elif nums[i]==0 and maxi[-1]:
maxi.append(0)
return max(maxi)
Although this code works & is efficient in terms of speed but in terms of space, it kinda sucks, because of the extra maxi array. So I gave it another shot, 2nd attempt (I wanted to reduce the space complexity, so I ditched the extra array and instead use just 2 variables):
def findMax( nums):
maxi = 0
current_maxi = 0
for i in range(len(nums)+1):
try:
if nums[i]:
current_maxi += 1
elif (nums[i]==0 and current_maxi>maxi):
maxi, current_maxi = current_maxi, 0
except IndexError:
if (current_maxi>maxi):
maxi, current_maxi = current_maxi, 0
return maxi
I've tested it out against so many custom test sets, but leetcode has some test cases on which it fails. This is a lot better than the 1st code in terms of speed and space, but where is it going wrong? can someone please help me know where this code goes wrong and is failing for some test cases?
spring grove
@fading sky does something like this work?
def consecutive_counter(array):
return np.diff(np.where(np.concatenate(([array[0]], array[:-1] != array[1:], [1])))[0])[::2].max()
havent really tested this but might give you an idea
there are probably also some solutions with itertools
or can you not use external libraries im unfamiliar with leetcode
topaz pulsar
Hello, I'm just getting started with leetcode and this was the first problem pre...
looking at the second function, what happens if nums[i] == 0 but current_maxi <= maxi?
I think this is where the issue lies
>>> findMax([1, 1, 1, 0, 1, 1, 0, 1, 1])
4```this is a case where currently your solution fails so it may be useful to use to help you fix it
idle cairn
i have one doubt about performance, i have a pickle file(the img) with a lot of dicts in rows and i need these lines to do query of names and take the language, u guys think if i use the languages names('en', 'fr', 'ru', 'es'...) as columns it could be faster? cause i know this way of keys as columns could be easier
How i gonna use this file -> example:
someone will type Carrot Seeds, so he will look at the column check if there is any Carrot Seeds in the values and if it has so he will take the language else he will say he dont found and suggest closest values
so what u guys think its the fast way to do it? using dicts as rows in a dataframe or using the languages as columns and say for he look at all collumns? (i'm using pandas to read the file)
fading sky
looking at the second function, what happens if `nums[i] == 0` but `current_maxi...
awesome, thankyou very much @topaz pulsar , but as you said when nums[i]==0 and current_maxi <= maxi, it isn't supposed to do anything because current_maxi <= maxi means that the current slice of 1s is not the longest chain of 1s, as there has already been a chain of 1s longer than the current one, which is what this condition current_maxi <= maxi generally means..?
topaz pulsar
well when you come across a 0 in the string of numbers, you want to setcurrent_maxi to 0 yes?, But looking at the line mentioned py elif (nums[i]==0 and current_maxi>maxi):will that always happen when a 0 is encountered in the list?
fading sky
Ahhhhh, GOT IT.
wow
WOW
I wrote a pretty shitty code then ig.
Thanks a ton @topaz pulsar π . you're awesome 
misty plume
hello I have a question, how would you get one element from a list that has 2 values per position?
ages_names = []
youngsters = []
names = []
for i in range(n):
name, age = map(str, input().strip().split())
names.append(name)
ages_names.append((int(age), name))
ages_names.sort()
for i in range(m):
youngsters.append(ages_names[i].)```I want to add only the names from the age_ names list
trim fiber
```n, m = map(int, input().strip().split())
ages_names = []
youngsters = []
nam...
When you have tuple then you can just use index notation
elements = [(0, "a"), (1, "b")]
print(elements[0][1]) # prints "a"
output: 5 2 JamesLineberger 55 JeanetteMaurey 73 ChristieDangelo 29 HeatherTrew 78 LeolaSwift 30 [(29, 'ChristieDangelo'), (30, 'LeolaSwift'), (55, 'JamesLineberger'), (73, 'JeanetteMaurey'), (78, 'HeatherTrew')] [(29, 'ChristieDangelo'), (30, 'LeolaSwift')]
output that I want: ChristieDangelo LeolaSwift
trim fiber
output: ```5 2
JamesLineberger 55
JeanetteMaurey 73
ChristieDangelo 29
HeatherTr...
Show your code with example data
misty plume
my code: ```
n, m = map(int, input().strip().split())
ages_names = []
youngsters = []
names = []
for i in range(n):
name, age = map(str, input().strip().split())
names.append(name)
ages_names.append((int(age), name))
ages_names.sort()
for i in range(m):
youngsters.append(ages_names[i])
print(ages_names)
print(youngsters)```
input: 5 2 JamesLineberger 55 JeanetteMaurey 73 ChristieDangelo 29 HeatherTrew 78 LeolaSwift 30
trim fiber
You are not using index here
for i in range(m):
youngsters.append(ages_names[i])
Like I said, you can use index to get values from tuple
!e
abc = (0, 1, 2)
print(abc[0], abc[1], abc[2])
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
0 1 2
You should use ages_names[i][1]
misty plume
that would give me only the name?
trim fiber
Try it π
trim fiber
It works?
misty plume
of course it does
trim fiber
π
trim fiber
Your welcome!
misty plume
legend x2
x4*
and now just to make yourself a god
how would I edit the output of the list
so that it doesn't have the brackets on the sides, and it's just the name with a space?
searched for it on google but didn't find it
trim fiber
how would I edit the output of the list
What do you mean? Do you want to pretty print a list?
trim fiber
!d str.join
halcyon plankBOT
trim fiber
!e
separator = "-,-"
collection = ["a", "b", "c", "d", "not so long", "I am very long element"]
print(separator.join(collection))
@trim fiber :white_check_mark: Your eval job has completed with return code 0.
a-,-b-,-c-,-d-,-not so long-,-I am very long element
For your case separator can be just " "
misty plume
the knowledge some people has amazes me up to the point of fear
halcyon plankBOT
Sorry, an unexpected error occurred. Please let us know!
DiscordServerError: 500 Internal Server Error (error code: 0): 500: Internal Server Error
misty plume
thanks :Β·3
trim fiber
the knowledge some people has amazes me up to the point of fear
Just few years of experience, keep learn and write code π
misty plume
oww, thank you, confidence boost included with the python code
topaz pulsar
another way you can do this (although this is probably more suited to when your elements aren't strings) is to use the sep kw arg of print, then you can unpack the list into print
>>> items = ["a", 123, True, ("my", "tuple")]
>>> print(*items)
a 123 True ('my', 'tuple')```(the standard value for `sep` is `" "`)
eternal mulch
I have a pandas dataframe with columns ['date', 'open', etc.]. How do i find the open on date "jan 01, 2019"?
eternal mulch
vocal gorge
I have a pandas dataframe with columns ['date', 'open', etc.]. How do i find the...
Filter the dataframe by a predicate:
df[df["date"] == "jan 01, 2019"]
that'll give you a slice of the dataframe containing all rows the date of which is this. Hopefully, that'll be one row. You can then take the ["open"] column of that slice, and the first (and hopefully only) element of it by [0].
vocal gorge
Wrong panda-related library; but sadly pandas has an unfun, not panda-related at all logo π
eternal mulch
f
nova owl
Hello Guys! I have a little problem on how to proceed.
I have a QTableWidget, when i double click on a cell I can modify things, but I really don't know how to get this action in the code in order to save the line or something. Do i need to "connect" somewhere ? Here is the relevant part of my code. Thank you very much for your time
class Ui_MainWindow(object):
def setupUi(self, MainWindow):
...
self.tableWidget = QtWidgets.QTableWidget(self.centralwidget)
self.tableWidget.setGeometry(QtCore.QRect(145, 51, 711, 501))
self.tableWidget.setObjectName("tableWidget")
self.tableWidget.setColumnCount(3)
self.tableWidget.setRowCount(0)
item = QtWidgets.QTableWidgetItem()
self.tableWidget.setHorizontalHeaderItem(0, item)
item = QtWidgets.QTableWidgetItem()
self.tableWidget.setHorizontalHeaderItem(1, item)
item = QtWidgets.QTableWidgetItem()
self.tableWidget.setHorizontalHeaderItem(2, item)
self.tableWidget.setColumnWidth(0,self.tableWidget.width()/3)
self.tableWidget.setColumnWidth(1,self.tableWidget.width()/3)
self.tableWidget.setColumnWidth(2,self.tableWidget.width()/3)
...
def retranslateUi(self, MainWindow):
_translate = QtCore.QCoreApplication.translate
MainWindow.setWindowTitle(_translate("MainWindow", "MainWindow"))
item = self.tableWidget.horizontalHeaderItem(0)
item.setText(_translate("MainWindow", "Ticker"))
item = self.tableWidget.horizontalHeaderItem(1)
item.setText(_translate("MainWindow", "Bottom"))
item = self.tableWidget.horizontalHeaderItem(2)
item.setText(_translate("MainWindow", "Top"))
def updateTable(self):
self.tableWidget.setRowCount(len(watchlist))
row = 0
for coin in watchlist:
self.tableWidget.setItem(row, 0, QtWidgets.QTableWidgetItem(coin.ticker))
self.tableWidget.setItem(row, 1, QtWidgets.QTableWidgetItem(str(coin.bottom)))
self.tableWidget.setItem(row, 2, QtWidgets.QTableWidgetItem(str(coin.top)))
row=row+1
Also, i believe i am in the right channel, I hesitated with user-interface
dapper sapphire
Is there an algorithm or pip module that could calculate how long execution of a program would take?
Or would doing end_time - start_time be the only way?
halcyon plankBOT
:incoming_envelope: :ok_hand: applied mute to @dire relic until 2021-04-19 23:46 (9 minutes and 59 seconds) (reason: duplicates rule: sent 4 duplicated messages in 10s).
vocal gorge
<@&831776746206265384>
stone sorrel
π
vocal gorge
Is there an algorithm or pip module that could calculate how long execution of a...
The builtin timeit, the somewhat nicer timerit package, but the most awesomest IMO are IPython's %time and %timeit line magic commands.
dapper sapphire
ok so let's say you wanted to create a progress bar for a program.
Then would you run the program and calculate the time using time or timeit and then hardcode that time to progress through the progress bar?
brave owl
im doing an easy leetcode problem and i came up with a solution with O(n) but its saying time limit exceeded?
how do i know how low i need to make the time complexity and how to even reduce it?
vocal gorge
ok so let's say you wanted to create a progress bar for a program.
Then would yo...
No, one would not use the time at all, but rather use some other metric you can track.
Say, if your code is a finite loop, track what percent of iterations have been done.
dapper sapphire
Oh yeah I see what you mean. Let's say if we are making 1,000 api calls. We would calculate the time it takes to make 20 calls and the time we get multiply that by 50, which would give us the estimated time it will take to make all 1,000 api calls
And we could extend the same thought process if we are iterating though a list that has 50,000 elements.
vocal gorge
pretty much.
Users mostly care about knowing what % is done
ETAs are infamously horribly inaccurate, so that's secondary π
dapper sapphire
ok that was helpful!! thanks a lot!
brave oak
how do i know how low i need to make the time complexity and how to even reduce ...
you don't, and it depends.
there could be a logarithmic or constant time solution
or linear time is the best you can do
but your constant is too high
or you're constantly hitting worst cases
etc.
brave owl
i mean its just an easy so i feel like linear should work
brave oak
i mean its just an easy so i feel like linear should work
show question and solution
brave owl
uh i figured it out lol
had an infinite loop
anyone have resources to continue learnintg ds&a?
i can do easy array/list problems and a few mediums but i have no idea what to do with other data structures
placid oasis
hello
keen hearth
i can do easy array/list problems and a few mediums but i have no idea what to d...
Check out the links pinned in this channel. I also recommend Khan Academy's algorithms course for an introduction to some of the theory.