[Numpy-discussion] flatten() without copy - is this possible?
dmitrey
openopt@ukr....
Tue Jun 5 12:06:18 CDT 2007
Thank you, but all your examples deal with 3-dimensional arrays. and I
still misunderstood, is it possible somehow for 2-dimensional arrays or no?
D.
Anne Archibald wrote:
> On 01/06/07, dmitrey <openopt@ukr.net> wrote:
>
>
>> y = x.flatten(1)
>>
>> turn array into vector (note that this forces a copy)
>>
>> Is there any way to do the trick wthout copying?
>> What are the problems here? Just other way of array elements indexing...
>>
>
> It is sometimes possible to flatten an array without copying and sometimes not.
>
> For numpy, a vector is a single block of memory in which there are
> elements of uniform type spaced at a uniform distance. This last is
> the key; it's called the "stride", and it need not be the same size as
> an element (so arange(10)[::3] can be created without a copy).
>
> A multidimensional array simply has many strides, one for each
> dimension. Thus ones((10,10,10)) simply keeps track of the stride for
> a row, the stride for a column, and the stride for a layer. If you
> want to transpose two axes, the data is not copied, instead the
> strides are simply exchanged. Under normal circumstances one need not
> care what the strides are or how the cells are laid out in memory as
> numpy hides that from normal users.
>
> What about flattening an array? It should turn an array into a vector,
> that is, take an array with n different strides and lengths and create
> as single array with a single stride and length. The order of the
> resulting elements needs to be specified; numpy normally defaults to
> "C order", which means that A[3,4,5] and A[3,4,6] are adjacent in the
> resulting array but A[3,4,5] and A[4,4,5] are not. (Note that this is
> a logical operation; the organization of the underlying array is
> irrelevant for the result.)
>
> If you want to ensure that no copy is made, you need to ensure that
> the stride between elements of the array you're flattening is always
> the same. Taking a 10-by-10-by-10 array A, the spacing between
> A[3,4,5] and A[3,4,6] needs to be the same as the spacing between
> A[3,4,6] and A[3,4,7]. This is automatic. But the spacing also needs
> to be the same as the spacing between A[3,4,9] and A[3,5,0]. This is
> not automatic, and often does not occur. In such cases numpy must make
> a copy to ensure that the resulting array is uniformly strided.
>
> What cases *don't* require a copy? Well, let's look at some examples:
>
> A = ones((10,10,10))
> reshape(A,(-1,)) # No copy needed
> reshape(A[:,:,:5],(-1,)) # Copy needed
> reshape(A[:,:,::2],(-1,)) # No copy needed
> reshape(A[:,::2,:],(-1,)) # Copy needed
> reshape(A[:5,:,:],(-1,)) # No copy needed
> reshape(A.transpose(),(-1,)) # Copy needed
>
> Note that none of the reindexing operations require a copy, but some
> of the reshapes do.
>
> It turns out to be nontrivial to detect all the cases where a copy can
> be avoided while reshaping, and IIRC numpy misses some (old versions
> of numpy almost always copied). But a freshly-created array is
> normally guaranteed to be reshapable without a copy.
>
> If you want to try reshaping an array without a copy, you can try
> assigning to .shape:
> In [3]: A = ones((10,10,10))[:,:5,:]
>
> In [4]: A.shape = (-1,)
> ---------------------------------------------------------------------------
> <type 'exceptions.AttributeError'> Traceback (most recent call last)
>
> /home/peridot/physics-projects/pulsed-flux/writings/<ipython console>
> in <module>()
>
> <type 'exceptions.AttributeError'>: incompatible shape for a
> non-contiguous array
>
> and
> In [7]: A = ones((10,10,10))[:5,:,:]
>
> In [8]: A.shape = (-1,)
>
>
> Anne
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>
>
>
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