Splitting list of strings based on a character in each string ( Python )

Question

So i have a list of strings that looks like this :

my_list = ['389.3K', '2M' , '1.9M' , '6.9M' , '4.3M' , '251.5K' , '3.6M']

This is how the list is generated and made readable ("video" is a list of selenium webelements ) :

my_list = [x.text for x in video]
video.extend(my_list)
my_list = [i for i in my_list if i if not 'ago' in i]
my_list = [w.replace("Views", "") for w in my_list]

What i want to do is SPLIT this list into two other lists based on ONE specific character in each element like so :

k_list = ['389.3K' , '251.5K']

m_list = ['2M' , '1.9M' , '6.9M' , '4.3M' , '3.6M']

My end goal is to be able to have only the numbers in the elements as a float and multiply each element by their appropriate amount ( K = *1000 and M = *1000000 ) like :

my_new_list = ['389,300' , '2,000,000‬' , '1,900,000' , '6,900,000‬' , '4,300,000', '251,500' , '3,600,000‬']

I'm new to python (coding in general tbh) so please excuse any spaghetti code or bad thought process.

This is what i tried :

k_val = "K"
m_val = "M"

if any(k_val in s for s in my_list):
    my_list = [w.replace("K", "") for w in my_list]
    my_list = [float(i) for i in vmy_list]
    my_list = [elem * 1000 for elem in my_list]
elif any(m_val in x for x in my_list):
    my_list = [w.replace("M", "") for w in my_lists]
    my_list = [float(i) for i in my_list]
    my_list = [elem * 1000000 for elem in my_list]

I get :

ValueError: could not convert string to float: '2M '

Unfortunately that didn't prevent some answerers from still solving Y... — superb rain
– superb rain, Commented Aug 10, 2020 at 12:03
My post was borked , i meant turn into float for the multiplication then back to string for further use. @superbrain — TheGreatPeanut
– TheGreatPeanut, Commented Aug 10, 2020 at 12:30

Rakesh · Accepted Answer · 2020-08-10 11:57:38Z

4

This is one approach

Ex:

my_list = ['389.3K', '2M' , '1.9M' , '6.9M' , '4.3M' , '251.5K' , '3.6M']
data = {"K": 1000, "M": 1000000}
result = [float(i[:-1])*data.get(i[-1], 0) for i in my_list]
print(result)

If you have multiple string in the end the use

import re
import locale    #https://stackoverflow.com/a/5180615/532312

locale.setlocale(locale.LC_ALL, '')

my_list = ['389.3K', '2M' , '1.9M' , '6.9M' , '4.3M' , '251.5K' , '3.6M']
data = {"K": 1000, "M": 1000000}

result = []
for i in my_list:
    m = re.match(r"(\d+\.?\d*)([A-Z])", i)
    if m:
        value, key = m.groups()
        result.append(locale.currency(float(value) * data.get(key, 0), symbol=False, grouping=True))
print(result)

Output:

['389,300.00', '2,000,000.00', '1,900,000.00', '6,900,000.00', '4,300,000.00', '251,500.00', '3,600,000.00']

edited Aug 10, 2020 at 11:57

answered Aug 10, 2020 at 11:51

Rakesh

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1 Comment

TheGreatPeanut Over a year ago

This works but i don't understand why yet. Thank you nevertheless and i will continue to learn until i understand why this actually works, i'll use the answer provided by @Daweo until then .

superb rain · Accepted Answer · 2020-08-10 11:57:43Z

1

If you actually want floats (you say you do, but then show strings):

>>> [float(s.replace('K', 'e3').replace('M', 'e6')) for s in my_list]
[389300.0, 2000000.0, 1900000.0, 6900000.0, 4300000.0, 251500.0, 3600000.0]

answered Aug 10, 2020 at 11:57

superb rain

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Comments

Daweo · Accepted Answer · 2020-08-10 12:00:13Z

0

As your input data are already repesenting float values I suggest to harness scientific-notation combined with float following way:

my_list = ['389.3K', '2M' , '1.9M' , '6.9M' , '4.3M' , '251.5K' , '3.6M']
e_list = [i.replace('K','e3').replace('M','e6') for i in my_list]
values = [float(i) for i in e_list]
my_new_list = [f'{int(i):,}' for i in values]
print(my_new_list)

Output:

['389,300', '2,000,000', '1,900,000', '6,900,000', '4,300,000', '251,500', '3,600,000']

answered Aug 10, 2020 at 12:00

Daweo

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Comments

stijndcl · Accepted Answer · 2020-08-10 11:56:56Z

0

You are getting this error, because in your first if you are only replacing the K's in case there's a K anywhere in the list, but that doesn't remove the M's (same goes for K's in the second if).

That way, you're trying to cast "2M" to a float because you only replaced the K's, but the M-terms still have their M's (and vice-versa). You should first create those two lists you mentioned, where you split them based on K and M, and then iterate through the K-list in the first if (and the M-list in the second if).

Creating those two separate lists can be done like this:

k_list = [val for val in my_list if k_val in val]
m_list = [val for val in my_list if m_val in val]

answered Aug 10, 2020 at 11:56

stijndcl

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1 Comment

TheGreatPeanut Over a year ago

I might try this is in my original spaghetti implementation. Thank you !!

Satish Michael · Accepted Answer · 2020-08-10 12:02:20Z

0

Here is another approach, probably not the best.

my_list = ['389.3K', '2M' , '1.9M' , '6.9M' , '4.3M' , '251.5K' , '3.6M']

k_list = [float(i[:-1])*1000 for i in my_list if i.endswith('K')]
m_list = [float(i[:-1])*1000000 for i in my_list if i.endswith('M')]


k_list_strings = [f'{num:,}' for num in k_list]
m_list_strings = [f'{num:,}' for num in m_list]

output

[389300.0, 251500.0]
[2000000.0, 1900000.0, 6900000.0, 4300000.0, 3600000.0]

output

['389,300.0', '251,500.0']
['2,000,000.0', '1,900,000.0', '6,900,000.0', '4,300,000.0', '3,600,000.0']

answered Aug 10, 2020 at 12:02

Satish Michael

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Comments

Ronald · Accepted Answer · 2020-08-10 12:12:56Z

0

A bit more explicit, and checking the input as well:

k_val = 'K'
m_val = 'M'
my_list = ['389.3K', '2M' , '1.9M' , '6.9M' , '4.3M' , '251.5K' , '3.6M']
kilos = []
megas = []
entire = []

for val in my_list:
    if val[-1] == k_val:
        fval = float(val[:-1]) * 1000
        kilos.append(fval)
    elif val[-1] == m_val:
        fval = float(val[:-1]) * 1000000
        megas.append(fval)
    else:
        print("detected invalid value: " + val)
        continue
    entire.append(fval)

print(str(kilos))
print(str(megas))
print(str(entire))

I like the approach of Rakesh and the others. But especially if one is new to programming, I like to be a little more verbose. Code golf is nice, but tends to be less easier to understand.

edited Aug 10, 2020 at 12:12

answered Aug 10, 2020 at 11:58

Ronald

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3 Comments

superb rain Over a year ago

How do you merge them back to their stated end goal?

Ronald Over a year ago

@superbrain Sorry, I missed that. It can be done in the loop (with the penalty of memory usage). If the kilos and megas aren't really needed, they can be eliminated (which then removes the penalty).

TheGreatPeanut Over a year ago

This answers my first questions and in a newbie friendly way!! Thank you. Also both Rakesh and @Daweo solved the overall problem.

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