🔒 list comprehension with dictionaries

[molten rune]

why cant i do this:

# list using dictionary
d = {'Name': 'John',
     'Grades': [100, 80, 20, 100]}
myList = [g for g in d[g]]

and yet this is a thing that works:

d = {'Name': 'John',
     'Grades': [100, 80, 20, 100]}
for g in d:
    print(d[g])

[signal pendant]

@molten rune

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[fathom sparrow]

how are u referring g without defining it

[molten rune]

g is just an iterable

[fathom sparrow]

!e ```py

list using dictionary

d = {'Name': 'John',
'Grades': [100, 80, 20, 100]}
myList = [d[g] for g in d]
print(myList)```

[molten rune]

it doesnt need to be defined

[signal pendant]

@fathom sparrow :white_check_mark: Your 3.12 eval job has completed with return code 0.

['John', [100, 80, 20, 100]]

[fathom sparrow]

it works now

!e d = {'Name': 'John',
'Grades': [100, 80, 20, 100]}
for g in d:
print(d[g])

[signal pendant]

@fathom sparrow :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | John
002 | [100, 80, 20, 100]

[fathom sparrow]

yup

[fathom sparrow]

wot

[molten rune]

ah so i need the d[g] at the start

[fathom sparrow]

it does

u r referring to it

without defining g

[molten rune]

d = {'Name': 'John',
     'Grades': [100, 80, 20, 100]}
for g in d:
    print(d[g])

g isnt defined here either

[turbid copper]

the loop defines g

there's more ways to assign data to a variable than just =

a for loop is one of those ways

[molten rune]

i dont really get why it works when u do d[g] for g in d

[turbid copper]

because that's the syntax of a list comprehension

[slender rover]

it will make your work done

[turbid copper]

<expression> for <variable> in <iterable>

[fathom sparrow]

what u wrote was

d = {'Name': 'John',
     'Grades': [100, 80, 20, 100]}
for g in d[g]:
    print(g)```

if we convert ur list comp literally

see where the issue is?

[molten rune]

yes i do in that loop

[slender rover]

no like what do you want to do ?

[turbid copper]

They're just trying to understand the syntax

the actual outcome here isn't important

[slender rover]

ohh

[turbid copper]

Very unhelpful.

[molten rune]

read the docs? lol...

[turbid copper]

They're here to discuss this with people

[molten rune]

so in the for loop the stuff on the left is the variable and the stuff on the right is the experssion?

[slender rover]

u mean they want to change the syntax by dicsussing it 🗿

[turbid copper]

the stuff indented below the for loop is the expression

[slender rover]

look do you know how the for loop works first ?

[fathom sparrow]

@slender rover if u arent interested in helping and only degrading OP, i suggest u stop msging instead

or go to a server that entertains degradation

[turbid copper]

green is the expression, blue is the variable, orange is the iterable

[slender rover]

wrong mention nvm i am trying to ! jst asking him if he knows for loop thing i would love to explain him the beauty of python.

[turbid copper]

You pinged the wrong user, but yes

[fathom sparrow]

oops 💀

[slender rover]

True

[molten rune]

ok this is immensily helpful lol

[turbid copper]

They'll ask if they need it

[slender rover]

Emotional Damage...

[turbid copper]

comprehension syntax can take a bit of getting used to since the expression comes first

[molten rune]

does it work the same way with dictionary comprehension?

[turbid copper]

yes except it allows for k:v syntax as the expression

[molten rune]

key:value right

[turbid copper]

!e

d = {'Name': 'John',
     'Grades': [100, 80, 20, 100]}

upper_dict = {k.upper():v for k, v in d.items()}
print(upper_dict)

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

{'NAME': 'John', 'GRADES': [100, 80, 20, 100]}

[turbid copper]

yes

this example will make all the keys uppercase strings

k.upper():v is the expression in this example

[molten rune]

can you use zip when iterating in a list comprehension?

[turbid copper]

d = {'Name': 'John',
     'Grades': [100, 80, 20, 100]}

upper_dict = {}

for k, v in d.items():
    upper_dict[k.upper()] = v

print(upper_dict)

[molten rune]

for like tuples

[turbid copper]

you can use whatever you want

if it works in a for loop, it will work in a list comp

[molten rune]

ok perfect

[molten rune]

can you use ternary operators inside our comprehensions?

[turbid copper]

yup!

[molten rune]

at the end?

[turbid copper]

No, it would be the expression

[molten rune]

or is that considered an expression?

ah

[turbid copper]

you can use just an else at the end of the list comp

it's a great way to filter things

[molten rune]

but if u wanna do ternary it has to be at the start

[turbid copper]

yes because it's an expression

!e

items = ['apples', 'bananas', 'apricots', 'cherries']

a_items = []

for item in items:
    if item[0] == 'a':
        a_items.append(item)

print(a_items)

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

['apples', 'apricots']

[turbid copper]

here's a basic for loop that filters out all items that don't begin with "a"

!e

items = ['apples', 'bananas', 'apricots', 'cherries']

a_items = [item for item in items if item[0] == 'a']
print(a_items)

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

['apples', 'apricots']

[turbid copper]

and here's the list comp equivalent

notice how the new list is shorter, because items have been filtered

if you were to use ternary, the resulting list would always have to be the same length

[molten rune]

so many items...

[turbid copper]

!e

items = ['apples', 'bananas', 'apricots', 'cherries']

a_items = [item if item[0] == 'a' else '<REMOVED>' for item in items]
print(a_items)

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

['apples', '<REMOVED>', 'apricots', '<REMOVED>']

[molten rune]

isnt the first index of a list the whole string

not just the first letter?

wouldnt item[0] be apple

[turbid copper]

no, items is the list. item is each individual string in the list

item[0] is the first letter of the string

it's common to use singular and plural variable names like this

plural for the list, singular for the individual item

[molten rune]

hmm ok wait but

when u iterate through a list arent u iterating through each full item in the list?

[turbid copper]

yes

for item in items

!e

items = ['apples', 'bananas', 'apricots', 'cherries']

for item in items:
    print(item[0])

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | a
002 | b
003 | a
004 | c

[turbid copper]

!e

items = ['apples', 'bananas', 'apricots', 'cherries']

print([item[0] for item in items])

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

['a', 'b', 'a', 'c']

[molten rune]

!e

list ['help', 'im', 'so', 'lost']
for i in list:
  print(list[i])

[signal pendant]

@molten rune :x: Your 3.12 eval job has completed with return code 1.

001 | Traceback (most recent call last):
002 |   File "/home/main.py", line 2, in <module>
003 |     for i in list:
004 | TypeError: 'type' object is not iterable

[slender rover]

[molten rune]

!e

list ['help', 'im', 'so', 'lost']
for i in list:
  print(i)

[turbid copper]

you don't really ever use [i] unless you're using range/enumerate

[signal pendant]

@molten rune :x: Your 3.12 eval job has completed with return code 1.

001 | Traceback (most recent call last):
002 |   File "/home/main.py", line 2, in <module>
003 |     for i in list:
004 | TypeError: 'type' object is not iterable

[turbid copper]

you don't have =

[molten rune]

omg

lol

yeah

i know

[turbid copper]

and I really recommend not naming a variable list

[molten rune]

T_T

!e

list1 = ['help', 'im', 'so', 'lost']
for i in list:
  print(list[i])

[signal pendant]

@molten rune :x: Your 3.12 eval job has completed with return code 1.

001 | Traceback (most recent call last):
002 |   File "/home/main.py", line 2, in <module>
003 |     for i in list:
004 | TypeError: 'type' object is not iterable

[turbid copper]

There's no rush when it comes to programming. Take your time to think about it

[molten rune]

ok im done

u get what im trying to do LOL

[turbid copper]

I don't

because that's invalid

[molten rune]

!e

list1 = ['help', 'im', 'so', 'lost']
for i in list:
  print(i)

[signal pendant]

@molten rune :x: Your 3.12 eval job has completed with return code 1.

001 | Traceback (most recent call last):
002 |   File "/home/main.py", line 2, in <module>
003 |     for i in list:
004 | TypeError: 'type' object is not iterable

[molten rune]

!e

list1 = ['help', 'im', 'so', 'lost']
for i in list1:
  print(i)

[signal pendant]

@molten rune :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | help
002 | im
003 | so
004 | lost

[molten rune]

oh

i get it now

[turbid copper]

that's basically no different from my last example

i is the elements of the list

not the index

[molten rune]

when u get the first indx of the iterable its the first letter

[turbid copper]

it's only the index when you use range or enumerate

it's not the index

which is even more of a reason I don't recommend using i as a variable unless you're dealing with indices

[slender rover]

i is basically a variable for the looping condition 😋 and list1 is the thing which it is looping in !

[turbid copper]

It's not a condition

it's a variable

[slender rover]

yes

corrected it ^^

[molten rune]

!e

list1 = ['help', 'im', 'so', 'lost']
for i in list1:
  print(i[0])

[signal pendant]

@molten rune :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | h
002 | i
003 | s
004 | l

[slender rover]

@molten rune see ! 😋

[turbid copper]

!e

list1 = ['help', 'im', 'so', 'lost']
for i in range(len(list1)):
  print(list1[i])

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | help
002 | im
003 | so
004 | lost

[turbid copper]

you're probably used to seeing it this way, using range

!e

list1 = ['help', 'im', 'so', 'lost']
for i in range(len(list1)):
  print(list1[i][0])

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | h
002 | i
003 | s
004 | l

[turbid copper]

In this case, we'd still need the [0] to access the first letter

[molten rune]

yeah i am lol

or with enumerate

[turbid copper]

yeah, no one will really use range(len(something)) professionally

unless you have a very specific reason

[molten rune]

when you refer to two things in like an x, y format

is that always referred to as zip?

[turbid copper]

no, zip() is a zip

x, y is "multiple assignment" or "unpacking"

zip returns something that is capable of being unpacked

[molten rune]

@turbid copper hey is there a way to iterate through 2 items in a list a time

so like if a list has 10 numbers u iterate over it 5 times

[turbid copper]

Sure, you could use a step/slice

[molten rune]

[::2] ?

like that?

[turbid copper]

This is one of those cases that I would use range/len

!e

letters = 'abcdefghij'

step_size = 2

for i in range(0, len(letters), step_size):
    print(letters[i:i+step_size])

[signal pendant]

@turbid copper :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | ab
002 | cd
003 | ef
004 | gh
005 | ij

[molten rune]

would enumerate work intead of range(len)?

[molten rune]

forgot to tag

[turbid copper]

no because enumerate doesn't have step

[molten rune]

is there a way to add items to an empty tuple?

[turbid copper]

No, tuples are immutable

[signal pendant]

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