Sort list of strings by a part of the string

You can use regex for this:

>>> import re
>>> r = re.compile(r'\((name\d+)\)')
>>> lis = ['variable1 (name1)', 'variable3 (name3)', 'variable2 (name100)']
>>> sorted(lis, key=lambda x:r.search(x).group(1))
['variable1 (name1)', 'variable2 (name100)', 'variable3 (name3)']

Note that above code will return something like name100 before name3, if that's not what you want then you need to do something like this:

>>> r = re.compile(r'\(name(\d+)\)')
def key_func(m):
    return int(r.search(m).group(1))

>>> sorted(lis, key=key_func)
['variable1 (name1)', 'variable3 (name3)', 'variable2 (name100)']

To change sorting key, use the key parameter:

>>>s = ['variable1 (name3)', 'variable2 (name2)', 'variable3 (name1)']
>>> s.sort(key = lambda x: x.split()[1])
>>> s
['variable3 (name1)', 'variable2 (name2)', 'variable1 (name3)']
>>> 

Works the same way with sorted:

>>>s = ['variable1 (name3)', 'variable2 (name2)', 'variable3 (name1)']
>>> sorted(s)
['variable1 (name3)', 'variable2 (name2)', 'variable3 (name1)']
>>> sorted(s, key = lambda x: x.split()[1])
['variable3 (name1)', 'variable2 (name2)', 'variable1 (name3)']
>>> 

Note that, as described in the question, this will be an alphabetical sort, thus for 2-digit components it will not interpret them as numbers, e.g. "11" will come before "2".

The solution is:

sorted(b, key = lambda x: x.split()[1])

Why? We want to sort the list (called b). As a key we will use (name X). Here we assume that it will be always preceded by space, therefore we split the item in the list to two and sort according to the second.