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Re: Slicing iterables in sub-generators without loosing elements

References <20120929161437.GA8832@taris.box>
From Ian Kelly <ian.g.kelly@gmail.com>
Date 2012-09-29 10:36 -0600
Subject Re: Slicing iterables in sub-generators without loosing elements
Newsgroups comp.lang.python
Message-ID <mailman.1638.1348936632.27098.python-list@python.org> (permalink)

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On Sat, Sep 29, 2012 at 10:14 AM, Thomas Bach
<thbach@students.uni-mainz.de> wrote:
> Hi,
>
> say we have the following:
>
>>>> data = [('foo', 1), ('foo', 2), ('bar', 3), ('bar', 2)]
>
> is there a way to code a function iter_in_blocks such that
>
>>>> result = [ list(block) for block in iter_in_blocks(data) ]
>
> evaluates to
>
>>>> result = [ [('foo', 1), ('foo', 2)], [('bar', 3), ('bar', 2)] ]
>
> by _only_ _iterating_ over the list (caching all the elements sharing
> the same first element doesn't count)?

Am I correct in understanding that the intent is that the "blocks" are
groups that share the same first item?

Is it guaranteed that the blocks will be contiguous?  If so, you could
use itertools.groupby:

from itertools import groupby, imap
from operator import itemgetter

def iter_in_blocks(data):
    return imap(itemgetter(1), groupby(data, itemgetter(0)))

If there is no such guarantee, then the list would need to be sorted first.

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Re: Slicing iterables in sub-generators without loosing elements Ian Kelly <ian.g.kelly@gmail.com> - 2012-09-29 10:36 -0600

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