jez jez - 9 months ago 53
Python Question

assigning to a wrapped slice of a numpy array

I have a large image

A
and a smaller image
B
, both expressed as 2-D
numpy
arrays. I want to use
A
as the canvas, and write translated copies of
B
all over it, packed in a hexagonal arrangement. The part I can't get my head around is how to handle it such that the image wraps both vertically and horizontally—essentially what I want is regular tessellation of a (padded, as necessary) sub-image onto a torus.

I've seen the discussion of
numpy.take
and
numpy.roll
at
wrapping around slices in Python / numpy and that shows me how to access and return a copy of a wrapped slice of an array, but I want to assign to that—i.e., for arbitrary integers
rowOffset
and
columnOffset
I want to do the equivalent of:

A = numpy.zeros((5,11), int)
B = numpy.array([[1,2,3,4,5,6,7]]) * numpy.array([[10,100,1000]]).T
# OK, we wouldn't be able to fit more than one or two copies of B into A, but they demonstrate the wrapped placement problem

wrappedRowIndices = ( numpy.arange(B.shape[0]) + rowOffset ) % A.shape[0]
wrappedColumnIndices = ( numpy.arange(B.shape[1]) + columnOffset ) % A.shape[1]
A[ wrappedRowIndices, : ][ :, wrappedColumnIndices ] = B


I see from a comment on the question,
and from a moment's reflection on the way
numpy
arrays are represented, that there's no way a wrapped slice can be returned as a
view
in the way this demands.

Is there (Y) a way of assigning to wrapped slices of an array in this way, or (X) an existing utility for performing the kind of tessellation I'm trying to achieve?

Answer Source

np.put is a 1d equivalent to np.take:

In [1270]: A=np.arange(10)
In [1271]: np.take(A,[8,9,10,11],mode='wrapped')
Out[1271]: array([8, 9, 0, 1])
In [1272]: np.put(A,[8,9,10,11],[10,11,12,13],mode='wrapped')
In [1273]: A
Out[1273]: array([12, 13,  2,  3,  4,  5,  6,  7, 10, 11])
In [1274]: np.take(A,[8,9,10,11],mode='wrapped')
Out[1274]: array([10, 11, 12, 13])

Its docs suggest np.place and np.putmask (and np.copyto). I haven't used those much, but it might be possible to construct a mask, and rearrangement of B that would do the copy.

=================

Here's an experiment with place:

In [1313]: A=np.arange(24).reshape(4,6)
In [1314]: mask=np.zeros(A.shape,bool)
In [1315]: mask[:3,:4]=True
In [1316]: B=-np.arange(12).reshape(3,4)

So I have mask the same size as A, with a 'hole' the size of B.

I can roll both the mask and B, and place the values in A in a wrapped fashion.

In [1317]: np.place(A, np.roll(mask,-2,0), np.roll(B,1,0).flat)
In [1318]: A
Out[1318]: 
array([[ -8,  -9, -10, -11,   4,   5],
       [  6,   7,   8,   9,  10,  11],
       [  0,  -1,  -2,  -3,  16,  17],
       [ -4,  -5,  -6,  -7,  22,  23]])

And with 2d rolls

In [1332]: m=np.roll(np.roll(mask,-2,0),-1,1)
In [1333]: m
Out[1333]: 
array([[ True,  True,  True, False, False,  True],
       [False, False, False, False, False, False],
       [ True,  True,  True, False, False,  True],
       [ True,  True,  True, False, False,  True]], dtype=bool)
In [1334]: b=np.roll(np.roll(B,1,0),-1,1)
In [1335]: b
Out[1335]: 
array([[ -9, -10, -11,  -8],
       [ -1,  -2,  -3,   0],
       [ -5,  -6,  -7,  -4]])
In [1336]: A=np.zeros((4,6),int)
In [1337]: np.place(A, m, b.flat)
In [1338]: A
Out[1338]: 
array([[ -9, -10, -11,   0,   0,  -8],
       [  0,   0,   0,   0,   0,   0],
       [ -1,  -2,  -3,   0,   0,   0],
       [ -5,  -6,  -7,   0,   0,  -4]])