Python：矩陣的非對角元素為0

Question

將方形對稱numpy ndarray的非對角線元素轉換為0的最快方法是什么？

Answer 1

我查看了保存對角線的速度，然后刪除矩陣，然后恢復對角線：

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

但是，如果矩陣不是很大，那么只需分配一個新矩陣就會更快

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

Answer 2

這是一個也適用於非連續數組的解決方案：

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]])