numpy-discussion

[Numpy-discussion] Compressed index sets, generators and iterators

[Bill B. <wb...@gm...>]

Has any thought been given to using compressed or functional representations
of index sets?

For instance there could be a where-like function that returns an object
that can generate a set of indexes on the fly, rather than explicitly
allocating arrays and enumerating all of the indices.

What's the reason iterators are not supported currently?
For instance  A[range(0,4)] works for a 1d A, but A[xrange(0,4)] does not.
Are iterators just too inefficient to bother with?

I could imagine an iterator that generates a set of N-tuples for an N-d
array being a legal indexing construct.  Even something like
A[idx_iterator] = value_iterator  would seem to make sense.

--bb

Re: [Numpy-discussion] Compressed index sets, generators and iterators

[A. M. A. <per...@gm...>]

On 01/11/06, Bill Baxter <wb...@gm...> wrote:

> What's the reason iterators are not supported currently?
> For instance  A[range(0,4)] works for a 1d A, but A[xrange(0,4)] does not.
> Are iterators just too inefficient to bother with?

If you want an iterator back, a generator comprehension will do it:
(A[i] for i in xrange(0,4))

If the result is to be a numpy array, the size must be known ahead of
time (so that the memory can be allocated) and specified. At this
point, and considering the overhead in calling back and forth to the
generator, the cost of converting to a list (after all,
A[list(xrange(0,40)] should work fine) isn't necessarily worth the
trouble.

A. M. Archibald

Re: [Numpy-discussion] Compressed index sets, generators and iterators

[Tim H. <tim...@ie...>]

A. M. Archibald wrote:
> On 01/11/06, Bill Baxter <wb...@gm...> wrote:
>
>   
>> What's the reason iterators are not supported currently?
>> For instance  A[range(0,4)] works for a 1d A, but A[xrange(0,4)] does not.
>> Are iterators just too inefficient to bother with?
>>     
>
> If you want an iterator back, a generator comprehension will do it:
> (A[i] for i in xrange(0,4))
>
> If the result is to be a numpy array, the size must be known ahead of
> time (so that the memory can be allocated) and specified. At this
> point, and considering the overhead in calling back and forth to the
> generator, the cost of converting to a list (after all,
> A[list(xrange(0,40)] should work fine) isn't necessarily worth the
> trouble.
>   
Another option for converting an iterator to a sequence that might be 
helpful, particularly if result is large, is to use fromiter. For example:

    a[numpy.fromiter(xrange(0,40, 2), int)]


-tim