使用掩码的逻辑索引适用于 numpy，不适用于 Matlab
Logical indexing using a mask works in numpy, not in Matlab
我正在尝试使用稀疏矩阵在 MATLAB 中重现以下 Python 代码。
>>> print(M)
[[0 0 0 0 0]
[0 1 1 1 0]
[0 1 0 1 0]
[0 1 1 1 0]
[0 0 0 0 0]]
>>> im2var = np.arange(5 * 5).reshape((5, 5))
>>> A = np.zeros((25, 25), dtype=int)
>>> A[im2var[M == 1], im2var[M == 1]] = 1
>>> print(A)
[[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
这是我用 MATLAB 写的
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
im2var = reshape(1:25, [5 5]);
A = zeros(25, 25);
A(im2var(M == 1), im2var(M == 1)) = 1;
num2str(A)
当我 运行 MATLAB 脚本时，我得到以下矩阵，这与 Numpy 输出明显不同。
ans =
25x73 char array
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
感谢您的帮助！
编辑：我也想完成以下效果，但目前的答案似乎不适用于两组索引。
在Python中：
>>> Mp = np.roll(M, 1, axis=1)
>>> A[im2var[M==1], im2var[Mp==1]] = -1
>>> print(A[5:15,5:15])
[[ 0 0 0 0 0 0 0 0 0 0]
[ 0 1 -1 0 0 0 0 0 0 0]
[ 0 0 1 -1 0 0 0 0 0 0]
[ 0 0 0 1 -1 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 1 -1 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 1 -1]
[ 0 0 0 0 0 0 0 0 0 0]]
根据当前答案的建议，我编写了以下代码。
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
Mp = circshift(M, 1, 2);
ind = find(M);
indp = find(Mp);
A = zeros(25, 25);
A(sub2ind(size(A), ind, ind)) = 1;
A(sub2ind(size(A), ind, indp)) = -1;
num2str(A)
虽然对角线的 1 已经成功出现，但 -1 出现在错误的位置。
ans =
25x73 char array
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
编辑 2：
根据编辑后的答案，我尝试了以下代码。
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
Mp = circshift(M, 1, 2);
ind = find(M);
indp = find(Mp.');
A = zeros(25, 25);
A(sub2ind(size(A), ind, ind)) = 1;
A(sub2ind(size(A), ind, indp)) = -1;
num2str(A(5:14, 5:14))
但它仍然不会产生与编辑 1 中的 Python 代码相同的结果。
ans =
10x28 char array
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 1 -1 0 0 0 0 0 0'
'0 0 0 1 -1 0 0 0 0 0'
'0 0 0 0 1 -1 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 1 -1 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 1'
在 MATLAB 中，您可以获得目标位置的 im2var(M == 1) 返回的 A 的相关行和列下标。这也可以用 find(M.') 来完成，而不需要初始化 im2var 或只是 find(M) 因为 M 在你的情况下等于 transpose(M) 。 find(M) returns 线性索引，其中 M 不为零，但 M 的线性索引与 A 的行和列下标相同。您不能直接使用这些行和列下标，需要将它们转换为线性索引，然后再进行，即
ind = find(M); % ind = find(M.'); in general
A(sub2ind(size(A),ind,ind)) = 1;
P.S: 请注意，MATLAB 遵循列主顺序，而 NumPy 遵循行主顺序。
我正在尝试使用稀疏矩阵在 MATLAB 中重现以下 Python 代码。
>>> print(M)
[[0 0 0 0 0]
[0 1 1 1 0]
[0 1 0 1 0]
[0 1 1 1 0]
[0 0 0 0 0]]
>>> im2var = np.arange(5 * 5).reshape((5, 5))
>>> A = np.zeros((25, 25), dtype=int)
>>> A[im2var[M == 1], im2var[M == 1]] = 1
>>> print(A)
[[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
这是我用 MATLAB 写的
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
im2var = reshape(1:25, [5 5]);
A = zeros(25, 25);
A(im2var(M == 1), im2var(M == 1)) = 1;
num2str(A)
当我 运行 MATLAB 脚本时，我得到以下矩阵，这与 Numpy 输出明显不同。
ans =
25x73 char array
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
感谢您的帮助！
编辑：我也想完成以下效果，但目前的答案似乎不适用于两组索引。
在Python中：
>>> Mp = np.roll(M, 1, axis=1)
>>> A[im2var[M==1], im2var[Mp==1]] = -1
>>> print(A[5:15,5:15])
[[ 0 0 0 0 0 0 0 0 0 0]
[ 0 1 -1 0 0 0 0 0 0 0]
[ 0 0 1 -1 0 0 0 0 0 0]
[ 0 0 0 1 -1 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 1 -1 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 1 -1]
[ 0 0 0 0 0 0 0 0 0 0]]
根据当前答案的建议，我编写了以下代码。
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
Mp = circshift(M, 1, 2);
ind = find(M);
indp = find(Mp);
A = zeros(25, 25);
A(sub2ind(size(A), ind, ind)) = 1;
A(sub2ind(size(A), ind, indp)) = -1;
num2str(A)
虽然对角线的 1 已经成功出现，但 -1 出现在错误的位置。
ans =
25x73 char array
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
编辑 2：
根据编辑后的答案，我尝试了以下代码。
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
Mp = circshift(M, 1, 2);
ind = find(M);
indp = find(Mp.');
A = zeros(25, 25);
A(sub2ind(size(A), ind, ind)) = 1;
A(sub2ind(size(A), ind, indp)) = -1;
num2str(A(5:14, 5:14))
但它仍然不会产生与编辑 1 中的 Python 代码相同的结果。
ans =
10x28 char array
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 1 -1 0 0 0 0 0 0'
'0 0 0 1 -1 0 0 0 0 0'
'0 0 0 0 1 -1 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 1 -1 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 1'
在 MATLAB 中，您可以获得目标位置的 im2var(M == 1) 返回的 A 的相关行和列下标。这也可以用 find(M.') 来完成，而不需要初始化 im2var 或只是 find(M) 因为 M 在你的情况下等于 transpose(M) 。 find(M) returns 线性索引，其中 M 不为零，但 M 的线性索引与 A 的行和列下标相同。您不能直接使用这些行和列下标，需要将它们转换为线性索引，然后再进行，即
ind = find(M); % ind = find(M.'); in general
A(sub2ind(size(A),ind,ind)) = 1;
P.S: 请注意，MATLAB 遵循列主顺序，而 NumPy 遵循行主顺序。