Em Sex, 2006-04-14 às 15:43 -0700, flamesrock escreveu: > Does anyone have a simple solution????
$ python2.4 Python 2.4.3 (#2, Mar 30 2006, 21:52:26) [GCC 4.0.3 (Debian 4.0.3-1)] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> l1 = [['c1',1],['c2',2],['c3',4]] >>> l2 = [['c1',1],['c2',2],['c4',4],['c3',3]] >>> print [tuple(x) for x in l1] [('c1', 1), ('c2', 2), ('c3', 4)] >>> print [tuple(x) for x in l2] [('c1', 1), ('c2', 2), ('c4', 4), ('c3', 3)] >>> s1 = frozenset(tuple(x) for x in l1) >>> s2 = frozenset(tuple(x) for x in l2) >>> print s1 frozenset([('c2', 2), ('c1', 1), ('c3', 4)]) >>> print s2 frozenset([('c2', 2), ('c4', 4), ('c3', 3), ('c1', 1)]) >>> print s2-s1 frozenset([('c4', 4), ('c3', 3)]) >>> print [list(x) for x in s2-s1] [['c4', 4], ['c3', 3]] -- Felipe. -- http://mail.python.org/mailman/listinfo/python-list