Chris Rebert wrote:
Ah, okay. Then you want:def all_same(lst): return len(set(lst)) == 1 def all_different(lst): return len(set(lst)) == len(lst) Note that these require all the elements of the list to be hashable.
This solution relies on the object ID -- no hashability required: # get list of object-IDs ids = map(lambda x: id(x), mylist) # ALL THE SAME? ... test whether "average ID" matches "first ID" sum(ids)/len(ids) == ids[0] -- http://mail.python.org/mailman/listinfo/python-list
