On Thu, 23 Apr 2009 21:51:42 -0400, Esmail wrote: >> set(a) == set(b) # test if a and b have the same elements >> >> # check that each list has the same number of each element # i.e. >> [1,2,1,2] == [1,1,2,2], but [1,2,2,2] != [1,1,1,2] for elem in set(a): >> a.count(elem) == b.count(elem) > > Ah .. this part would take care of different number of duplicates in the > lists. Cool.
At significant cost of extra work. Counting the number of times a single element occurs in the list is O(N). Counting the number of times every element occurs in the list is O(N**2). Sorting is O(N*log N), so for large lists, sorting will probably be much cheaper. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
