🔒 List formatting problem I can't solve (beginner programmer)

[turbid wadi]

hi guys, super beginner programmer so noob question, I have a list of strings alternating numbers and numbers+" bpm"(eg: ['123123','123123 bpm','123123','1231231 bpm']) I am trying to remove all the " bpm" from the odd-indexed strings then convert every element of the list from a string into a float when there's only numbers left. However my brain is about to explode and I'm not making any progress, Can someone help me with this code?
Code:

indexCounter = 0
for el in list:
    if indexCounter%2==1:
        list[indexCounter]=list[el[:-4]]
    list[indexCounter]=float(el)
    indexCounter += 1

I am returned with TypeError: list indices must be integers or slices, not str

[dusky summit]

@turbid wadi

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[severe vault]

you could loop over the list, then for each element, check if the element contains the string "bpm", and if it does, extract it out using string slicing from 0 to the index of "bpm", remove all trailing whitespace with the string.strip() method, and convert it into a float, otherwise convert the element into a float directly

if you want to modify the original list you might need to use a range-based for loop

!e

elements = ["1", "1 bpm", "2", "2 bpm"]
for i in range(len(elements)):
    element = elements[i]
    if "bpm" in element:
        elements[i] = float(element[0:element.index("bpm")].strip())
    else:
        elements[i] = float(element)
print(elements)

[dusky summit]

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

[1.0, 1.0, 2.0, 2.0]

[feral rock]

if this is a beginner problem i am less than a stick

[turbid wadi]

can u also remove the white space by adding the space to the truncate? like

elements[i] = float(element[0:element.index(" bpm")])

[severe vault]

sure, if your strings containing "bpm" are always of the form "100 bpm" and not "100bpm"

also not "100 bpm"

[turbid wadi]

also could you explain why the ranged-based for loop is necessary? compared to just a normal for loop? is it just to have the index counter integrated?

[severe vault]

if you're looking to modify the list you're iterating over then i don't think you can use a for element in list loop

if you're making a new list then you can use that kind of loop

[turbid wadi]

I see..

[ivory terrace]

every for loop in python is a ranged-based for loop. it does not have normal for loops

[severe vault]

difference is you're iterating over a range instead of the list itself

[turbid wadi]

!e

elements = ["1", "1 bpm", "2", "2 bpm, 2, 2 bpm"]
for i in range(len(elements)):
    element = elements[i]
    if "bpm" in element:
        elements[i] = float(element[0:element.index("bpm")].strip())
    else:
        elements[i] = float(element)
print(elements)

[dusky summit]

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

[1.0, 1.0, 2.0, 2.0]

[severe vault]

usually you don't want to modify the object you're iterating over

[turbid wadi]

oof, duplicates are an issue

I'm assuming the duplicates dont work because when it calls elements[i] it calls the first accessed only?

so in that case, should I use a separate counter instead that goes up after every iteration to forcibly call the next in the index?

[severe vault]

not sure what you're looking for, you want to remove duplicates?

[turbid wadi]

no I wanna keep duplicates

but still iterate through them too

[severe vault]

i'm sorry i don't really understand

[turbid wadi]

I added some extra elements to the list on this one

but it output only 4 elements instead of the 6 in the list

so I was assuming that it didnt work for duplicates

oh wait no did IU just not type in the quotations? wait lemme try that again

!e

elements = ["1", "1 bpm", "2", "2 bpm", "2", "2 bpm"]
for i in range(len(elements)):
    element = elements[i]
    if "bpm" in element:
        elements[i] = float(element[0:element.index("bpm")].strip())
    else:
        elements[i] = float(element)
print(elements)

[dusky summit]

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

[1.0, 1.0, 2.0, 2.0, 2.0, 2.0]

[turbid wadi]

okay nvm

I'm jsut slow

😭

[severe vault]

oh i see now, if there's multiple occurrences of bpm you'll have to somehow remove all of them while bpm exists in the element

like while "bpm" in element:

and then slice using the index of each "bpm"

[turbid wadi]

no I don't think there will be any multiple instances in one string, I just messe dup testing it

haha

[severe vault]

alright cool

[normal tide]

!e

elements = ["1", "1 bpm", "2", "2 bpm", "2", "2 bpm"]

print([float(element.split()[0]) for element in elements])

[dusky summit]

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

[1.0, 1.0, 2.0, 2.0, 2.0, 2.0]

[severe vault]

why didn't i think of split...

[normal tide]

That should work for 1 or more white spaces I think

[turbid wadi]

is this just taking the first number of each element?

[normal tide]

.split() splits a string on white space by deafult, in this list comphrension we're just taking the first element which happens to be our numbers

[severe vault]

it's splitting each part of the string separated by a space in the list into a temporary list, then takes the first element of that temporary list, which is just the number

[normal tide]

!e

print([float(ele.split()[0]) for ele in ['123123','123123 bpm','123123','1231231 bpm']])

[dusky summit]

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

[123123.0, 123123.0, 123123.0, 1231231.0]

[turbid wadi]

so to store it back in the original list it'd be this?

elements[element] = [float(element.split()[0]) for element in elements]
print(elements)

!e

elements[element] = [float(element.split()[0]) for element in elements]
print(elements)

[dusky summit]

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

001 | Traceback (most recent call last):
002 |   File "/home/main.py", line 1, in <module>
003 |     elements[element] = [float(element.split()[0]) for element in elements]
004 |                                                                   ^^^^^^^^
005 | NameError: name 'elements' is not defined. Did you mean: 'element'?

[turbid wadi]

oh forgot to put the list in

!e

elements = ["1", "1 bpm", "2", "2 bpm", "2", "2 bpm"]
elements[element] = [float(element.split()[0]) for element in elements]
print(elements)

[dusky summit]

: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 |     elements[element] = [float(element.split()[0]) for element in elements]
004 |              ^^^^^^^
005 | NameError: name 'element' is not defined. Did you mean: 'elements'?

[turbid wadi]

I-

[normal tide]

It's a really good idea to just create a brand new list instead of trying to update or overwrite the old one. This is true for a lot of objects because mutable things are very hard to keep track of compared to dealing with things as if they are immutable

[turbid wadi]

ohh it can't read tthat because I haven't initialized element yet

and I'm calling it

[turbid wadi]

so

!e

elements = ["1", "1 bpm", "2", "2 bpm", "2", "2 bpm"]
newElements = [float(element.split()[0]) for element in elements]
print(elements)

[dusky summit]

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

['1', '1 bpm', '2', '2 bpm', '2', '2 bpm']

[turbid wadi]

!e

elements = ["1", "1 bpm", "2", "2 bpm", "2", "2 bpm"]
newElements = [float(element.split()[0]) for element in elements]
print(newElements)

[dusky summit]

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

[1.0, 1.0, 2.0, 2.0, 2.0, 2.0]

[turbid wadi]

like that?

[normal tide]

Yes exactly, try and use snake_case since this is Python and give it a good enough name so you know what you've got there and that's perfect

!e

bpms = ['123123','123123 bpm','123123','1231231 bpm']

processed_bpms = [float(ele.split()[0]) for ele in bpms]

print(processed_bpms)

[dusky summit]

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

[123123.0, 123123.0, 123123.0, 1231231.0]

[turbid wadi]

my greatest weakness, readability 😭 pep8 is so hard

[normal tide]

Just try to keep your future self in mind when naming things, all devs spend more time reading code then they do writing it so make it easier to read every chance you get

[turbid wadi]

okay, so followup question

I now need to group them in 2s and write it out into a txt file where inside it would just be a long string of characters like this: 67.12|69,90|1500 with every element grouped with other element. so for that example the original string would have been:

['69','67.12 bpm','90','1500 bpm']

oh god

I just realized this small program I thought would take me 30 mins

is a lot harder than I expected

WHY IS THE ORDER THE OTHER WAY AROUND

omg

so I'm assuming, iterate through list, split odd from even indices, and then put them back except the odd ones first instead of the even ones to get a list with the right order, but now I just need to know how to write it out into a txt file

as a long string with "|" in the middle of each instead of commas

[ivory terrace]

pep8 has nothing to do with readability, it's just a style that guido van rossum likes, you can also use camel case if you like it better, the main thing is to keep a consistent style in your project

(or it has to do with readability but but that doesn't mean that other styles aren't just as readable)

[normal tide]

There is an easy enough way to take pairs of elements from a list, let me see if I can recall it

def pair_off(lst):
    """
    Takes a 1D list and returns a list of tuples of the items in the list paired off.

    Parameters:
    lst (list): The input list.

    Returns:
    list: A list of tuples with paired items.
    """
    return [(lst[i], lst[i + 1]) for i in range(0, len(lst) - 1, 2)]

[turbid wadi]

hold on so

def paired_list(lst):
    return [(lst [i+1], lst[i]) for i in range(0, len(lst) - 1,2)]

ultraProcessedBPMS = paired_list(processed_bpms)
print(ultraProcessedBPMS)

so like that?

[normal tide]

!e

def pair_off(lst):
    return [(lst[i], lst[i + 1]) for i in range(0, len(lst) - 1, 2)]

def clean_bpms(bpms):
    return [float(ele.split()[0]) for ele in bpms]

bpms = ['69','67.12 bpm','90','1500 bpm']

strings_to_concat = [f"{pair[0]}|{pair[1]}" for pair in pair_off(clean_bpms(bpms))]

strings_to_concat

final_string = ",".join(strings_to_concat)

print(final_string)

[dusky summit]

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

69.0|67.12,90.0|1500.0

[normal tide]

yeah something like paired_bpms would probably be fine if you know the flow of your code. Once names get too long they also become hard to read and understand

There might also be a csv writer that Python has to help with writing this list of strings to a file versus creating one big string to write.

If you want to be really efficient there is always converting the list comphrensions into generators instead but I don't know what scale you're working on.

[turbid wadi]

so just end it with something like this?

with open("BPMOutput.txt", "w") as text_file:
    text_file.write(final_string)

[normal tide]

Yeah that could work depending on the scale, there are other ways to write it if you know it's going to be a csv

# Example data
header = ['Name', 'Age', 'City']
data = [
    ['Alice', 30, 'New York'],
    ['Bob', 25, 'Los Angeles'],
    ['Charlie', 35, 'Chicago']
]

with open('output.csv', 'w', newline='') as file:
    writer = csv.writer(file)
    writer.writerow(header)  # Write the header
    writer.writerows(data)   # Write the data

If your project is small enough you can probably hold the whole string in memory and just go about writing it that way, if you're dealing with a LOT of data then using generators and the csv writer to write rows as they become available is probably the very best way to do it.

You'll have to decide what you need for your project

[turbid wadi]

OMG IT WORKS

YOU GUYS ARE GODS AMONGST MEN

THANK YOU

💕

!close

[dusky summit]

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