[Tutor] List methods inside a dictionary of list

[Rafael Turner]

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 <swiftone at swiftone.org> wrote:
> On Mon, Jul 11, 2011 at 9:26 AM, Rafael Turner
> <steven.rafael.turner at 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
> swiftone at swiftone.org
>