Michael M. a écrit : > How to find the longst element list of lists? For what definition of "find" ? You want the lenght of the longest sublist, it's index, or a reference to it ?
> I think, there should be an easier way then this: > > s1 = ["q", "e", "d"] > s2 = ["a", "b"] > s3 = ["a", "b", "c", "d"] Err... this makes three distinct lists, not a list of lists. > if len(s1) >= len(s2) and len(s1) >= len(s3): > sx1=s1 ## s1 ist längster > if len(s2) >= len(s3): > sx2=s2 > sx3=s3 > else: > sx2=s3 > sx3=s2 > (snip repeated code) Looks like it would be time to learn how to factor out repetitions... > After, the list ist sorted: > > sx1 = ["a", "b", "c", "d"] > sx2 = ["q", "e", "d"] > sx3 = ["a", "b"] > This is still not a list of lists. Now for the answer, sorted() is your friend: print sorted([s1, s2, s3], key=list.__len__, reverse=True) => [['a', 'b', 'c', 'd'], ['q', 'e', 'd'], ['a', 'b']] # Or if you really want sx1, sx2 and sx3: sx1, sx2, sx3 = sorted([s1, s2, s3], key=list.__len__, reverse=True) Is that easier enough ?-) -- http://mail.python.org/mailman/listinfo/python-list
