John Posner <jjpos...@snet.net> writes:
> # 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]
I don't think you can rely on id's being the same if what
you want is that the values are the same:
>>> a = "foo" + "bar"
>>> b = "foobar"
>>> a==b
True
>>> id(a) == id(b)
False
I'd use:
from operator import eq
all_the_same = reduce(eq, mylist)
I don't see how to do all_different in less than quadratic time,
without using hashing or sorting:
all_different = all(sum(1 for y in mylist if x==y)==1 for x in mylist)
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