On Mon, Jul 11, 2011 at 9:26 AM, Rafael Turner <steven.rafael.tur...@gmail.com> wrote: > I am playing lists and dictionaries and I came across this > counter-intuitive result. > >>>> d = dict(zip(['a', 'q', 'c', 'b', 'e', 'd', 'g', 'j'],8*[[0]])) ... >>>> d['a'].__setitem__(0,4) ... > > I was not expecting all the keys to be updated. Is there any > documentation I could read on how different datatypes' methods and > operators interact differently when inside a dictionary? I would also > like to find a way of being able to use list methods in side a > dictionary so that
As has been mentioned, this isn't the dictionary doing anything weird, this is that "8*[[0]]" gives you a list of 8 references to the same list. You can play with just that part and see that that's the source of your issue. To achieve what you are trying, try this instead: d = dict([(x,[0]) for x in ['a', 'q', 'c', 'b', 'e', 'd', 'g', 'j']]) Can you understand how this behaves differently than 8*[[0]] ? Check the Python docs for array multiplication if you're confused, but the basic idea is that "[0]" isn't getting evaluated freshly for every piece in the array for 8*[[0]], but in a list comprehension it is. -- Brett Ritter / SwiftOne swift...@swiftone.org _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
