Scott David Daniels wrote: > Dan Sommers wrote: >> ... >> longest_list, longest_length = list_of_lists[ 0 ], len( longest_list >> ) for a_list in list_of_lists[ 1 : ]: >> a_length = len( a_list ) >> if a_length > longest_length: >> longest_list, longest_length = a_list, a_length >> will run faster than sorting the list just to pick off one element (O(n) >> vs. O(n log n) for all of you Big-Oh notation fans out there; you know >> who you are!). > > Or, more succinctly, after: > list_of_lists = [["q", "e", "d"], > ["a", "b"], > ["a", "b", "c", "d"]] > You can find the longest with: > maxlength, maxlist = max((len(lst), lst) for lst in list_of_lists) > or (for those pre-2.5 people):
pre-2.4 > maxlength, maxlist = max([(len(lst), lst) for lst in list_of_lists]) With the caveat that if there are lists of equal length their items will be compared: >>> max((len(lst), lst) for lst in [[1], [1j]]) Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: no ordering relation is defined for complex numbers So if you want the first out of the lists with equal length on a python where Steve's enhancement is not yet available (pre-2.5): >>> max_len, dummy, max_list = max((len(lst), -i, lst) for i, lst in enumerate([[1], [1j]])) >>> max_len, max_list (1, [1]) Peter -- http://mail.python.org/mailman/listinfo/python-list
