poster
2019-12-10, 16:49
说我有以下矩阵：
A = randi(10, [6 3]) 7 10 3 5 5 7 10 5 1 6 5 10 4 9 1 4 10 1 我想提取每两行并将它们放入第三维，所以结果将是：
B(:,:,1) = 7 10 3 5 5 7 B(:,:,2) = 10 5 1 6 5 10 B(:,:,3) = 4 9 1 4 10 1 我显然可以使用for循环来做到这一点，只是想知道如何使用permute / reshape / ..作为单线更优雅地做到这一点（注意矩阵大小和步长必须是参数）
% params step = 5; r = 15; c = 3; % data A = randi(10, [rc]); B = zeros(step, c, r/step); % assuming step evenly divides r % fill counter = 1; for i=1:step:r B(:,:,counter) = A(i:i+step-1, :); counter = counter + 1; end
回答：
这是使用reshape (https://www.mathworks.com/help/matlab/ref/reshape.html)和permute (https://www.mathworks.com/help/matlab/ref/permute.html)的单行解决方案：
C = 3; % Number of columns R = 6; % Number of rows newR = 2; % New number of rows A = randi(10, [RC]); % 6-by-3 array of random integers B = permute(reshape(A.', [C newR R/newR]), [2 1 3]); 当然，这要求newR被R newR 。
更多&回答... (https://stackoverflow.com/questions/1390909)
A = randi(10, [6 3]) 7 10 3 5 5 7 10 5 1 6 5 10 4 9 1 4 10 1 我想提取每两行并将它们放入第三维，所以结果将是：
B(:,:,1) = 7 10 3 5 5 7 B(:,:,2) = 10 5 1 6 5 10 B(:,:,3) = 4 9 1 4 10 1 我显然可以使用for循环来做到这一点，只是想知道如何使用permute / reshape / ..作为单线更优雅地做到这一点（注意矩阵大小和步长必须是参数）
% params step = 5; r = 15; c = 3; % data A = randi(10, [rc]); B = zeros(step, c, r/step); % assuming step evenly divides r % fill counter = 1; for i=1:step:r B(:,:,counter) = A(i:i+step-1, :); counter = counter + 1; end
回答：
这是使用reshape (https://www.mathworks.com/help/matlab/ref/reshape.html)和permute (https://www.mathworks.com/help/matlab/ref/permute.html)的单行解决方案：
C = 3; % Number of columns R = 6; % Number of rows newR = 2; % New number of rows A = randi(10, [RC]); % 6-by-3 array of random integers B = permute(reshape(A.', [C newR R/newR]), [2 1 3]); 当然，这要求newR被R newR 。
更多&回答... (https://stackoverflow.com/questions/1390909)