使用掩码的逻辑索引在numpy中工作,而不是在Matlab中

2024-06-07 10:47:46 发布

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我试图用一个稀疏矩阵在MATLAB中重现下面的Python代码。你知道吗

>>> 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,我得到了下面的矩阵,它与Numpy输出明显不同。你知道吗

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'

谢谢你的帮助!你知道吗

编辑:我也想达到以下效果,但目前的答案似乎不适用于两组索引。你知道吗

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]]

通过使用当前答案的建议,我编写了以下代码。你知道吗

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1已经成功地出现了,但是-1在错误的位置。你知道吗

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。你知道吗

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))

但是它仍然不能产生与edit1中的Python代码相同的结果。你知道吗

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'

Tags: 代码sizenpzerosmpfindprintmatlab
1条回答
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1楼 · 发布于 2024-06-07 10:47:46

在MATLAB中,可以得到目标位置的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,而无需初始化im2var或仅仅find(M),因为在您的例子中M等于transpose(M)find(M)返回线性索引,其中M不是零,但M的线性索引与A的行和列下标相同。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,然后继续

ind = find(M);     % ind = find(M.'); in general
A(sub2ind(size(A),ind,ind)) = 1;

p.S:请注意,MATLAB遵循列的主要顺序,而NumPy遵循行的主要顺序。你知道吗

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