I did not understand the behavior of array multiplication. In fact, I just now learned what it was called thanks to your email.
Best wishes, Rafael On Mon, Jul 11, 2011 at 9:33 AM, Brett Ritter <swift...@swiftone.org> wrote: > 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
