On Fri, 11 Mar 2011 13:00:23 -0800, Patrick wrote:
> Hi,
>
> The build-in map functions looks quite nice but it requests the
> iterables to be of the same length or otherwises will file with None
> (most of time fails the function). Just wondering if there are already
> enhancement work done with it?
That has been fixed in Python 3.1:
>>> from operator import add
>>> list(map(add, [1,2,3], [1,2,3,4,5]))
[2, 4, 6]
Starting from Python 2.3, the itertools module has had a function imap
with the same behaviour:
>>> from operator import add
>>> list(itertools.imap(add, [1,2,3], [1,2,3,4,5]))
[2, 4, 6]
> I did some simple code but it will handle list without nesting only.
Surely that is a good thing. If you have a mix of nested and non-nested
data, that tells me your data is badly organized and needs to be fixed.
> I am looking for examples that could hand input of "a = [2,3], b=4"
Using a small helper function:
import itertools
def make_iterable(obj):
try:
iter(obj)
except TypeError:
obj = itertools.cycle([obj])
return obj
def enhanced_map(func, *args):
args = map(make_iterable, args)
return list(itertools.imap(func, *args))
WARNING: this is dangerous. If none of the arguments are iterable, e.g.
you call enhanced_map(operator.add, 2, 3), this will happily loop forever
building a larger and larger list, until you run out of memory or kill
the Python process.
> and "a=[1,[2,3],4], b=[5,[6,7,8],9,10]". That means if the nesting
> structure is the same, enhanced map function will automatically extend
> the shorter list using the last element.
It isn't clear what you want here. Are you expecting this enhanced map to
recursively drop down into each layer of sub-sequences? That is:
enhanced_map([1, [2,3, [4,5], 6], 7], [8, [7,6, [5,4], 3], 2])
should be the same as
map([1, 2, 3, 4, 5, 6, 7], [8, 7, 6, 5, 4, 3, 2])
or something different?
What do you expect to happen if the sub-sequences don't match up exactly?
E.g. a = [1, 2, [3, 4]]; b = [1, [2, 3], 4]
What do you expect to happen if the shorter list is empty?
E.g. a = [1, 2, [3, 4], 5]; b = [1, 2, [], 3]
This will get really messy fast. My advice is to forget about this as a
bad idea, and instead concentrate on making sure your data isn't so badly
arranged in the first place.
--
Steven
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