June 3, 2008
7:48 p.m.
From: "Arnaud Delobelle" <arnodel@googlemail.com>
there no way to predict the output of this:
set_of_sets = { {1, 2}, {2, 3} } set.difference(*set_of_sets)
That's silly. Lot's of functions do odd things with random argument ordering:
s = set([9, 3]) int.__sub__(*s) 6
Besides, you can already run the sample fragment in Py2.5:
sos = set( [frozenset([1, 2]), frozenset([2, 3])]) frozenset.difference(*sos) frozenset([1])
Raymond