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
我想提取每个2行,放到第三个层面,所以结果会是这样:
B(:,:,1) =
7 10 3
5 5 7
B(:,:,2) =
10 5 1
6 5 10
B(:,:,3) =
4 9 1
4 10 1
我可以明显地做到这一点有一个for循环,只是不知道如何使用作为一个班轮置换 / 重塑做到这一点更优雅/ ..(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
下面是一个使用一个在线解决方案RESHAPE和置换 :
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]);
当然,这需要newR
整除R
这里有一个一行与reshape
和permute
,但没有调换输入数组-
out = permute(reshape(A,newR,size(A,1)/newR,[]),[1 3 2]);
其中newR
处于行数3D
阵列输出。
标杆
本节建议的形式给出在这个岗位比较反对other solution with reshape, permute & transpose
性能。 该datasizes充气比例在问题中列出的那些。 因此, A
是60000 x 300
大小,我们将它分开,使得3D
输出将具有200 rows
,因此变暗-3将具有300
的条目。
基准测试代码 -
%// 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.