New submission from Kirk McDonald: One question which is asked with surprising frequency in #python is how to yield multiple objects at a time from an iterable object. That is, given [1, 2, 3, 4, 5, 6], get [(1, 2), (3, 4), (5, 6)].
The "grouper" function in the itertools recipes page provides one
pattern for this. A similar function (which behaves differently when the
length of the iterable is not evenly divisible by n) looks like this:
def group(iterable, n=2):
return itertools.izip(*(iter(iterable),)*n)
This code is fairly opaque to the novice. It is ugly, and takes a bit of
head-scratching to realize exactly what it is doing. Because this
operation is asked for with some frequency, and because this general
implementation is so ugly, I believe it belongs in the library.
There is a related function which is asked for much less frequently, but
which is a more general case of the group() function listed above. This
other function has a third "step" argument. This argument controls how
far each group is in advance of the previous group. For example:
list(group([1, 2, 3, 4], n=2, step=1)) -> [(1, 2), (2, 3), (3, 4)]
The original function is equivalent to this function when step is equal
to n.
Please find attached a patch with implementation, documentation, and tests.
----------
components: Library (Lib)
files: itertools.group.patch
messages: 58716
nosy: KirkMcDonald
severity: normal
status: open
title: Add group() to itertools
versions: Python 2.6
Added file: http://bugs.python.org/file8979/itertools.group.patch
__________________________________
Tracker <[EMAIL PROTECTED]>
<http://bugs.python.org/issue1643>
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itertools.group.patch
Description: Binary data
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