I have a tensor A such that A.shape = (32, 19, 2) and a binary matrix B such that B.shape = (32, 19). Is there a one-line operation I can perform to get a matrix C, where C.shape = (32, 19) and C(i,j) = A[i, j, B[i,j]]?
Essentially, I want to use B as an indexing matrix, where if B[i,j] = 1 I take A[i,j,1] to form C(i,j).
np.where to the rescue. It's the same principle as mtrw's answer:
In [344]: A=np.arange(4*3*2).reshape(4,3,2)
In [345]: B=np.zeros((4,3),dtype=int)
In [346]: B[[0,1,1,2,3],[0,0,1,2,2]]=1
In [347]: B
Out[347]:
array([[1, 0, 0],
[1, 1, 0],
[0, 0, 1],
[0, 0, 1]])
In [348]: np.where(B,A[:,:,1],A[:,:,0])
Out[348]:
array([[ 1, 2, 4],
[ 7, 9, 10],
[12, 14, 17],
[18, 20, 23]])
np.choose can be used if the last dimension is larger than 2 (but smaller than 32). (choose operates on a list or the 1st dimension, hence the rollaxis.
In [360]: np.choose(B,np.rollaxis(A,2))
Out[360]:
array([[ 1, 2, 4],
[ 7, 9, 10],
[12, 14, 17],
[18, 20, 23]])
B can also be used directly as an index. The trick is to specify the other dimensions in a way that broadcasts to the same shape.
In [373]: A[np.arange(A.shape[0])[:,None], np.arange(A.shape[1])[None,:], B]
Out[373]:
array([[ 1, 2, 4],
[ 7, 9, 10],
[12, 14, 17],
[18, 20, 23]])
This last approach can be modified to work when B does not match the 1st 2 dimensions of A.
np.ix_ may simplify this indexing
I, J = np.ix_(np.arange(4),np.arange(3))
A[I, J, B]
You can do it using list comprehension:
C = np.array([[A[i, j, B[i, j]] for j in range(A.shape[1])] for i in range(A.shape[0])])
C = A[:,:,0]*(B==0) + A[:,:,1]*(B==1) should work. You can generalize this as np.sum([A[:,:,k]*(B==k) for k in np.arange(A.shape[-1])], axis=0) if you need to index more planes.
