I have this list:

list = ['12kg','900g', '2,7kg', '801g', '15', '3,63kg', '1kg]

and I want to remove only letter g (or replace with "") when I have pattern <number>g, which means:

new_list = ['12kg', '900', '2,7kg', '801', '15', '3,63kg', '1kg]

I've tried to write the following but failed they all failed:

• [^kK]g: for some reason gets last digit (0g and 1g)
• [^kK]g$: nothing comes up (I expected the sign $ would get the end of string)
• [^kK](g): it does work but in a group -> How do I replace the group match with what I want?

CodePudding user response：

Is this what you're looking for?

import re


lst = ['12kg', '900g', '2,7kg', '801g', '15', '3,63kg', '1kg']
newLst = [re.sub('([0-9] )g$', lambda x: x.groups()[0], el) for el in lst]
print(newLst)  # => ['12kg', '900', '2,7kg', '801', '15', '3,63kg', '1kg']

The regex I used here was ([0-9] )g$. This will match the string with at least one number and end with "g". The re.sub call will then replace the string with only the number part.

CodePudding user response：

How about:

import re
list = ['12kg','900g', '2,7kg', '801g', '15', '3,63kg', '1kg']
[re.sub('(\d)g', r'\1',x) for x in list]

['12kg', '900', '2,7kg', '801', '15', '3,63kg', '1kg']

This matches a digit followed by g, and substitutes back the digit alone. The trick is \d matches a digit, (\d) puts the matched digit into a group, and \1 substitute group 1, which is the digit without the g.

CodePudding user response：

If you know the "g" is always at the end, you don't need a regex:

new_list = [x[:-1] for x in list]

p.s. Don't use the name list in your own code since there is already a list in the Python builtin package.