On Mar 13, 6:34 pm, Noah <[EMAIL PROTECTED]> wrote:
> What is the fastest way to select N items at a time from a dictionary?
> I'm iterating over a dictionary of many thousands of items.
> I want to operate on only 100 items at a time.
> I want to avoid copying items using any sort of slicing.
> Does itertools copy items?
>
> This works, but is ugly:
>
> >>> from itertools import *
> >>> D = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6, 'g':7, 'h':8, 'i':9,
> >>> 'j':10}
> >>> N = 3
> >>> for G in izip(*[chain(D.items(), repeat(None, N-1))]*N):
>
> ... print G
> ...
> (('a', 1), ('c', 3), ('b', 2))
> (('e', 5), ('d', 4), ('g', 7))
> (('f', 6), ('i', 9), ('h', 8))
> (('j', 10), None, None)
>
> I'd prefer the last sequence not return None
> elements and instead just return (('j',10)), but this isn't a huge
> deal.
>
> This works and is clear, but it makes copies of items:
>
> >>> ii = D.items()
> >>> for i in range (0, len(ii), N):
>
> ... print ii[i:i+N]
> ...
> [('a', 1), ('c', 3), ('b', 2)]
> [('e', 5), ('d', 4), ('g', 7)]
> [('f', 6), ('i', 9), ('h', 8)]
> [('j', 10)]
>
groupby?
import itertools
D = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6, 'g':7, 'h':8, 'i':9,
'j':10}
N = 3
it = itertools.groupby(enumerate(D.items()), lambda t: int(t[0]/N))
for each in it:
print tuple(t[1] for t in each[1])
--
Hope this helps,
Steven
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