opposite of zip()?

[Steven D'Aprano]

On Fri, 14 Dec 2007 21:47:06 -0800, igor.tatarinov wrote:

> Given a bunch of arrays, if I want to create tuples, there is
> zip(arrays). What if I want to do the opposite: break a tuple up and
> append the values to given arrays:
>    map(append, arrays, tupl)
> except there is no unbound append() (List.append() does not exist,
> right?).


Don't guess, test.

>>> list.append  # Does this exist?
<method 'append' of 'list' objects>


Apparently it does. Here's how *not* to use it to do what you want:

>>> arrays = [[1, 2, 3, 4], [101, 102, 103, 104]]
>>> tupl = tuple("ab")
>>> map(lambda alist, x: alist.append(x), arrays, tupl)
[None, None]
>>> arrays
[[1, 2, 3, 4, 'a'], [101, 102, 103, 104, 'b']]

It works, but is confusing and hard to understand, and the lambda 
probably makes it slow. Don't do it that way.



> Without append(), I am forced to write a (slow) explicit loop:
>   for (a, v) in zip(arrays, tupl):
>       a.append(v)

Are you sure it's slow? Compared to what?


For the record, here's the explicit loop:

>>> arrays = [[1, 2, 3, 4], [101, 102, 103, 104]]
>>> tupl = tuple("ab")
>>> zip(arrays, tupl)
[([1, 2, 3, 4], 'a'), ([101, 102, 103, 104], 'b')]
>>> for (a, v) in zip(arrays, tupl):
...     a.append(v)
...
>>> arrays
[[1, 2, 3, 4, 'a'], [101, 102, 103, 104, 'b']]


I think you're making it too complicated. Why use zip()?


>>> arrays = [[1, 2, 3, 4], [101, 102, 103, 104]]
>>> tupl = tuple("ab")
>>> for i, alist in enumerate(arrays):
...     alist.append(tupl[i])
...
>>> arrays
[[1, 2, 3, 4, 'a'], [101, 102, 103, 104, 'b']]




-- 
Steven