Sorting a list of strings based on numeric order of numeric part

You could use the re module to split each string into a tuple of characters and grouping the digits into one single element. Something like r'(\d+)|(.)'. The good news with this regex is that it will return separately the numeric and non numeric groups.

As a simple key, we could use:

def key(x):
    # the tuple comparison will ensure that numbers come before letters
    return [(j, int(i)) if i != '' else (j, i)
        for i, j in re.findall(r'(\d+)|(.)', x)]

Demo:

lst = ['a1a', 'a2a', 'a5b', 'a10a', 'b1a', 'abc']
print(sorted(lst, key=key)

gives:

['a1a', 'a2a', 'a5b', 'a10a', 'abc', 'b1a']

If you want a more efficient processing, we could compile the regex only once in a closure

def build_key():
    rx = re.compile(r'(\d+)|(.)')
    def key(x):
        return [(j, int(i)) if i != '' else (j, i)
            for i, j in rx.findall(x)]
    return key

and use it that way:

sorted(lst, key=build_key())

giving of course the same output.