[Numpy-discussion] Interleaved Arrays and
Neil Martinsen-Burrell
nmb@wartburg....
Tue Jun 16 21:22:40 CDT 2009
On 2009-06-16 16:05 , Robert wrote:
> Neil Martinsen-Burrell wrote:
>> On 06/16/2009 02:18 PM, Robert wrote:
>>> >>> n = 10
>>> >>> xx = np.ones(n)
>>> >>> yy = np.arange(n)
>>> >>> aa = np.column_stack((xx,yy))
>>> >>> bb = np.column_stack((xx+1,yy))
>>> >>> aa
>>> array([[ 1., 0.],
>>> [ 1., 1.],
>>> [ 1., 2.],
>>> [ 1., 3.],
>>> [ 1., 4.],
>>> [ 1., 5.],
>>> [ 1., 6.],
>>> [ 1., 7.],
>>> [ 1., 8.],
>>> [ 1., 9.]])
>>> >>> bb
>>> array([[ 2., 0.],
>>> [ 2., 1.],
>>> [ 2., 2.],
>>> [ 2., 3.],
>>> [ 2., 4.],
>>> [ 2., 5.],
>>> [ 2., 6.],
>>> [ 2., 7.],
>>> [ 2., 8.],
>>> [ 2., 9.]])
>>> >>> np.column_stack((aa,bb))
>>> array([[ 1., 0., 2., 0.],
>>> [ 1., 1., 2., 1.],
>>> [ 1., 2., 2., 2.],
>>> [ 1., 3., 2., 3.],
>>> [ 1., 4., 2., 4.],
>>> [ 1., 5., 2., 5.],
>>> [ 1., 6., 2., 6.],
>>> [ 1., 7., 2., 7.],
>>> [ 1., 8., 2., 8.],
>>> [ 1., 9., 2., 9.]])
>>> >>> cc = _
>>> >>> cc.reshape((n*2,2))
>>> array([[ 1., 0.],
>>> [ 2., 0.],
>>> [ 1., 1.],
>>> [ 2., 1.],
>>> [ 1., 2.],
>>> [ 2., 2.],
>>> [ 1., 3.],
>>> [ 2., 3.],
>>> [ 1., 4.],
>>> [ 2., 4.],
>>> [ 1., 5.],
>>> [ 2., 5.],
>>> [ 1., 6.],
>>> [ 2., 6.],
>>> [ 1., 7.],
>>> [ 2., 7.],
>>> [ 1., 8.],
>>> [ 2., 8.],
>>> [ 1., 9.],
>>> [ 2., 9.]])
>>> >>>
>>>
>>>
>>> However I feel too, there is a intuitive abbrev function like
>>> 'interleave' or so missing in numpy shape_base or so.
>>
>> Using fancy indexing, you can set strided portions of an array equal to
>> another array. So::
>>
>> In [2]: aa = np.empty((10,2))
>>
>> In [3]: aa[:, 0] = 1
>>
>> In [4]: aa[:,1] = np.arange(10)
>>
>> In [5]: bb = np.empty((10,2))
>>
>> In [6]: bb[:,0] = 2
>>
>> In [7]: bb[:,1] = aa[:,1] # this works
>>
>> In [8]: cc = np.empty((20,2))
>>
>> In [9]: cc[::2,:] = aa
>>
>> In [10]: cc[1::2,:] = bb
>>
>> In [11]: cc
>> Out[11]:
>> array([[ 1., 0.],
>> [ 2., 0.],
>> [ 1., 1.],
>> [ 2., 1.],
>> [ 1., 2.],
>> [ 2., 2.],
>> [ 1., 3.],
>> [ 2., 3.],
>> [ 1., 4.],
>> [ 2., 4.],
>> [ 1., 5.],
>> [ 2., 5.],
>> [ 1., 6.],
>> [ 2., 6.],
>> [ 1., 7.],
>> [ 2., 7.],
>> [ 1., 8.],
>> [ 2., 8.],
>> [ 1., 9.],
>> [ 2., 9.]])
>>
>> Using this syntax, interleave could be a one-liner.
>>
>> -Neil
>
> that method of 'filling an empty with a pattern' was mentioned in
> the other (general) interleaving question. It requires however a
> lot of particular numbers and :'s in the code, and requires even
> more statements which can hardly be written in functional style -
> in one line?. The other approach is more jount, free of fancy
> indexing assignments.
jount? I think that assigning to a strided index is very clear, but
that is a difference of opinion. All of the calls to np.empty are the
equivalent of the column_stack's in your example. I think that
operations on segments of arrays are fundamental to an array-processing
language such as NumPy. Using ";" you can put as many of those
statements as you would like one line. :)
> The general interleaving should work efficiently in one like this:
>
> np.column_stack/concatenate((r,g,b,....), axis=...).reshape(..)
>
> But as all this is not intuitive, something like this should be in
> numpy perhaps? :
>
> def interleave( tup_arrays, axis = None )
Here is a minimally tested implementation. If anyone really wants this
for numpy, I'll gladly add comments and tests. I couldn't figure out
how to automatically find the greatest dtype, so I added an argument to
specify, otherwise it uses the type of the first array.
def interleave(arrays, axis=0, dtype=None):
assert len(arrays) > 0
first = arrays[0]
assert all([arr.shape == first.shape for arr in arrays])
new_shape = list(first.shape)
new_shape[axis] *= len(arrays)
if dtype is None:
new_dtype = first.dtype
else:
new_dtype = dtype
interleaved = np.empty(new_shape, new_dtype)
axis_slice = [slice(None, None, None)]*axis + \
[slice(0,None,len(arrays))] + [Ellipsis]
for i, arr in enumerate(arrays):
axis_slice[axis] = slice(i, None, len(arrays))
interleaved[tuple(axis_slice)] = arr
return interleaved
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