On Wed, 15 Apr 2009 23:56:45 +0100, John Posner <jjpos...@snet.net> wrote:
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]
Oh John! After all the recent discussion of identity versus equality too.
my_list = [ 1024 ] my_list.append(1024) # Defeat the interpreter's cunningness ids = map(lambda x: id(x), ml) ids
[9864656, 9864704]
sum(ids)/len(ids) == ids[0]
False -- Rhodri James *-* Wildebeeste Herder to the Masses -- http://mail.python.org/mailman/listinfo/python-list
