[Tutor] itertools.izip question?
Kent Johnson
kent37 at tds.net
Mon Oct 15 17:41:38 CEST 2007
Iyer wrote:
>
>
> I have a dictionary
>
> d = {0: array([0, 0], dtype=uint16), 1: array([1, 1], dtype=uint16), 2:
> array([2, 2], dtype=uint16), 3: array([3, 3], dtype=uint16)}
> or
> d = {0:d0,1:d1,2:d2,3:d3}
>
>
> d0, d1, d2 and d3 are numpy.ndarray type
>
> I wish to get the interleaved data by using
>
> outputdata = numpy.array([item for items in itertools.izip(d.values())
> for item in items])
Try itertools.izip(*d.values())
What you are doing is passing a single list to izip(), just as if you
said itertools.izip([d0,d1,d2,d3]). (Note the extra brackets compared to
your sample below.)
The *d.values() means, use this list as the parameter list (instead of
as a single parameter).
Kent
>
> But I get outputdata = array([[0, 0], [1, 1], [2, 2],
> [3, 3]], dtype=uint16)
>
> which is different from the desired answer, which can be obtained by
>
> outputdata = numpy.array([item for items in itertools.izip(d0,d1,d2,d3)
> for item in items])
>
> outputdata = array([0, 1, 2, 3, 0, 1, 2, 3], dtype=uint16)
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