Python：矩阵的非对角元素为0

Question

将方形对称numpy ndarray的非对角元素转换为0的最快方法是什么？

Answer 1

我检查了保存对角线的速度，然后切断矩阵，然后恢复对角线：

n = len(mat)
d = mat.ravel()[::n+1]
values = d.copy()
mat[:,:] = 0
d[:] = values

如果矩阵不是很大，则可能只是分配一个新的更快

mat = numpy.diag(numpy.diag(mat))

Answer 2

这是一个也适用于非连续数组的解决方案：

a = np.arange(110).reshape(10, 11)[:, :10]

diag = np.einsum('ii->i', a)
# or if a is not guaranteed to be square
# mn = min(a.shape)
# diag = np.einsum('ii->i', a[:mn, :mn])
save = diag.copy()
a[...] = 0
diag[...] = save

a

# array([[  0,   0,   0,   0,   0,   0,   0,   0,   0,   0],
#        [  0,  12,   0,   0,   0,   0,   0,   0,   0,   0],
#        [  0,   0,  24,   0,   0,   0,   0,   0,   0,   0],
#        [  0,   0,   0,  36,   0,   0,   0,   0,   0,   0],
#        [  0,   0,   0,   0,  48,   0,   0,   0,   0,   0],
#        [  0,   0,   0,   0,   0,  60,   0,   0,   0,   0],
#        [  0,   0,   0,   0,   0,   0,  72,   0,   0,   0],
#        [  0,   0,   0,   0,   0,   0,   0,  84,   0,   0],
#        [  0,   0,   0,   0,   0,   0,   0,   0,  96,   0],
#        [  0,   0,   0,   0,   0,   0,   0,   0,   0, 108]])