问题
I would like to construct an array with the following structure:
A[i,j,i,j,] = B[i,j]
with all other entries 0: A[i,j,l,k]=0 # (i,j) =\= (l,k)
I.e. if I have the B
matrix constructed how can I create the matrix A
, preferably in a vectorized manner.
Explicitly, let B = [[1,2],[3,4]]
Then:
A[1,1,:,:]=[[1,0],[0,0]]
A[1,2,:,:]=[[0,2],[0,0]]
A[2,1,:,:]=[[0,0],[3,0]]
A[2,2,:,:]=[[0,0],[0,4]]
回答1:
We can use an open grid to assign to A
broadcasting the indexing arrays across the axes:
B = np.array([[1,2],[3,4]])
i,j = B.shape
A = np.zeros([i,j,i,j])
i, j = np.ogrid[:i, :j]
A[i,j,i,j] = B
print(A)
array([[[[1., 0.],
[0., 0.]],
[[0., 2.],
[0., 0.]]],
[[[0., 0.],
[3., 0.]],
[[0., 0.],
[0., 4.]]]])
回答2:
This is my solution with indexing:
x,y = np.meshgrid(np.arange(B.shape[1]),np.arange(B.shape[0]))
x,y = x.ravel(), y.ravel()
A = np.zeros(B.shape + B.shape)
A[y.ravel(), x.ravel(), y.ravel(), x.ravel()] = B.ravel()
# checking
for i in range(2):
for j in range(2):
print(f'A[{i},{j}]:',A[i,j])
Output:
A[0,0]: [[1. 0.]
[0. 0.]]
A[0,1]: [[0. 2.]
[0. 0.]]
A[1,0]: [[0. 0.]
[3. 0.]]
A[1,1]: [[0. 0.]
[0. 4.]]
来源:https://stackoverflow.com/questions/61826786/numpy-symmetric-4d-matrix-construction