Zipping a dictionary whose values are lists
Peter Otten
__peter__ at web.de
Thu Apr 12 12:42:38 EDT 2012
tkpmep at gmail.com wrote:
> I using Python 3.2 and have a dictionary
>>>> d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
>
> whose values are lists I would like to zip into a list of tuples. If I
> explicitly write:
>>>> list(zip([1,2], [1,2,3], [1,2,3,4])
> [(1, 1, 1), (2, 2, 2)]
>
> I get exactly what I want. On the other hand, I have tried
>
>>>>list(zip(d))
> [(0,), (1,), (2,)]
>
>>>> list(zip(d.values()))
> [([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
>
>>>> list(zip(d[i] for i in d))
> [([1, 2],), ([1, 2, 3],), ([1, 2, 3, 4],)]
>
>>>> list(zip(*d))
> Traceback (most recent call last):
> File "<pyshell#48>", line 1, in <module>
> list(zip(*d))
> TypeError: zip argument #1 must support iteration
>
> and nothing quite works. What am I doing wrong?
You have all the building blocks ;)
>>> d = {0:[1,2], 1:[1,2,3], 2:[1,2,3,4]}
>>> list(zip(*d.values()))
[(1, 1, 1), (2, 2, 2)]
The order of the values is undefined, so you may want to sort the lists by
key first:
>>> list(zip(*[v for k, v in sorted(d.items())]))
[(1, 1, 1), (2, 2, 2)]
Well, I guess it doesn't really matter for that example...
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