如何在Python中使用掩码对数组的数组进行卷积？

Question

Given a t1xt2xn array and a m1xm2 mask, how to obtain the t1xt2xn array where the n-dim arrays are convolved with the mask?

The function scipy.signal.convolve is not able to handle this because it only accept inputs with same number of dimensions.

Example with the "same" logic:

in1 =
[[[0,1,2],[3,4,5],[6,7,8]],
[[9,10,11],[12,13,14],[15,16,17]],
[[18,19,20],[21,22,23],[24,25,26]]]

in2 =
[[0,1,0],
[0,1,0],
[0,1,0]]

output =
[[[0,0,0],[15,17,19],[0,0,0]],
[[0,0,0],[36,39,42],[0,0,0]],
[[0,0,0],[33,35,37],[0,0,0]]]

Answer 1

I'm very sorry but I haven't strong math background, so my answer could be wrong. Anyway, if you need to use mask for selecting you should convert it to bool type. For example:

in1 = np.array([[[0,1,2],    [3,4,5],    [6,7,8]],
                [[9,10,11],  [12,13,14], [15,16,17]],
                [[18,19,20], [21,22,23], [24,25,26]]])

in2 = np.array([[0, 1, 0],
                [0, 1, 0],
                [0, 1, 0]])

mask = in2.astype(bool)

print(in1[mask])
# [[ 3  4  5]
#  [12 13 14]
#  [21 22 23]]

in3 = np.zeros(in1.shape)

in3[mask] = np.convolve(in1[mask].ravel(), in2.ravel(), 'same').reshape(mask.shape)

print(in3)

# [[[  0.   0.   0.]
#   [ 15.  17.  19.]
#   [  0.   0.   0.]]
# 
#  [[  0.   0.   0.]
#   [ 36.  39.  42.]
#   [  0.   0.   0.]]
# 
#  [[  0.   0.   0.]
#   [ 33.  35.  37.]
#   [  0.   0.   0.]]]

I'm not very sure about last part, especially about reshaping but I hope you get an idea.