On 31 October 2012 23:09, Steven D'Aprano <steve+comp.lang.pyt...@pearwood.info> wrote: > The trick is to take each string and split it into a leading number and a > trailing alphanumeric string. Either part may be "empty". Here's a pure > Python solution: > > from sys import maxsize # use maxint in Python 2 > def split(s): > for i, c in enumerate(s): > if not c.isdigit(): > break > else: # aligned with the FOR, not the IF > return (int(s), '') > return (int(s[:i] or maxsize), s[i:]) > > Now sort using this as a key function: > > py> L = ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] > py> sorted(L, key=split) > ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2']
You don't actually need to split the string, it's enough to return a pair consisting of the number of leading digits followed by the string as the key. Here's an implementation using takewhile: >>> from itertools import takewhile >>> def prefix(s): ... return sum(1 for c in takewhile(str.isdigit, s)) or 1000, s ... >>> L = ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] >>> sorted(L, key=prefix) ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2'] Here's why it works: >>> map(prefix, L) [(1, '9'), (4, '1000'), (1000, 'abc2'), (2, '55'), (1, '1'), (1000, 'abc'), (2, '55a'), (1, '1a')] -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
