将长2D矩阵分割成第三维

Question

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，因此结果如下：

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循环执行此操作，只是想知道如何使用permute / reshape / ..更加优雅地进行单线程 （注意矩阵大小和步骤必须是参数）

% params
step = 5;
r = 15;
c = 3;

% data
A = randi(10, [r c]);
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

Answer 1

这是使用reshape和permute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, [R C]);  % 6-by-3 array of random integers
B = permute(reshape(A.', [C newR R/newR]), [2 1 3]);

这当然要求newR均匀地划分为R

Answer 2

这是一个带有reshape和permute ，但没有转置输入数组 -

out = permute(reshape(A,newR,size(A,1)/newR,[]),[1 3 2]);

，其中newR是3D数组输出中的行数。

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标杆

本节将此帖中提出的other solution with reshape, permute & transpose并对表现other solution with reshape, permute & transpose 。 数据集与问题中列出的数据集成比例。 因此， A是60000 x 300大小，我们将它拆分，使得3D输出将具有200 rows ，因此dim-3将具有300个条目。

基准代码 -

%// Input
A = randi(10, [60000 300]); %// 2D matrix
newR = 200;                 %// New number of rows

%// Warm up tic/toc.
for k = 1:50000
    tic(); elapsed = toc();
end

N_iter = 5; %// Number of iterations for each approach to run with

disp('---------------------- With PERMUTE, RESHAPE & TRANSPOSE')
tic
for iter = 1:N_iter
    [R,C] = size(A);
    B = permute(reshape(A',[C newR R/newR]),[2 1 3]); %//'
end
toc, clear B R C iter

disp('---------------------- With PERMUTE & RESHAPE')
tic
for iter = 1:N_iter
    out = permute(reshape(A,newR,size(A,1)/newR,[]),[1 3 2]);
end
toc

输出 -

---------------------- With PERMUTE, RESHAPE & TRANSPOSE
Elapsed time is 2.236350 seconds.
---------------------- With PERMUTE & RESHAPE
Elapsed time is 1.049184 seconds.