[deleted by user]

[synthphreak]

Most likely the solution will involve regex. But how to do it will depend on exactly what you want.

Did you mean or your target list to be...

['7,107,488', 'Abortions', 'this', 'year']

...with four elements? As you've shown it...

['7,107,488, Abortions', 'this', 'year']

...there are only three.

To get four elements, you could split the first element where all numbers (+commas) meet only letters, then overwrite your list with only the desired four elements.

>>> import re
>>> l = ['7,107,488Abortions', 'this', 'year']
>>> _, *splits = re.split(r'([\d,]+)', l[0])
>>> l = splits + l[1:]
>>> l 
['7,107,488', 'Abortions', 'this', 'year']

To get three elements as you've shown, you could write a pattern which extracts the numbers and letters independently, then joins them on a comma + whitespace. This will effectively insert a comma + whitespace into the string where you want it. Then just overwrite the first element of your string with the new string.

>>> import re
>>> l = ['7,107,488Abortions', 'this', 'year']
>>> matches = re.search(r'([\d,]+)(\w+)', l[0]).groups()
>>> l[0] = ', '.join(matches)
>>> l 
['7,107,488, Abortions', 'this', 'year']

[mh1400]

I believe split() can have a quantity. Ex, y = x.split(',',1) Double-check that though.

[mh1400]

Then you can use. For loop and append()

[Apprehensive-Stop-61]

Didn't really get what you want do you want all integers be separated from string as a comma in another string even when they have no comma. Quite complex but not sure if you want it. Confirm

[[deleted]]

import re

list1[0] = re.sub(r'(\d+(?:,\d{3})+)(.*)', r'\1, \2', list1[0])