Question

我希望获得一个矩阵，该矩阵由矩阵中的每个3x3正方形组成。矩阵是＃通道乘以m×n。最终结果是一个大小为#channels x (size1-2)*(size2-2) x 9的矩阵，我可以通过以下代码以非向量的方式进行此操作：

clear
size1 = 10;
size2 = 10;
matrix = 1:size1*size2;
matrix =reshape(matrix,[size1 size2]);
matrix_withdraw = 1:(88*size1*size2);
matrix_withdraw = reshape(matrix_withdraw,[88 size1 size2]);
tic
iter = 1;
for ii = 1:size1-2
    for jj = 1:size2-2
        locs(ii,jj,:,:) = matrix(ii:ii+2,jj:jj+2);
        method1(:,iter,:) = reshape(matrix_withdraw(:,ii:ii+2,jj:jj+2),[88,9]);
        iter = iter+1;
    end
end
locs = permute(locs,[3 4 1 2]);

对于大尺寸的矩阵，这显然相当慢。我正在寻求向量化。我设法获得了一个可行的解决方案，但它不是很干净

locs2 = ones(size1*(size2-2),1)+(0:size1*(size2-2)-1)';
temp = 1:size1*(size2-2);
temp = mod(temp,size1);
temp = (temp>(size1-2)) | (temp==0);
locs2(temp,:) = [];
locs3 = reshape(locs2,[1 1 (size1-2) (size2-2)]);
locs3(2,1,:,:) = reshape(locs2+1,[1 1 (size1-2) (size2-2)]);
locs3(3,1,:,:) = reshape(locs2+2,[1 1 (size1-2) (size2-2)]);
locs3(:,2,:,:) = locs3(:,1,:,:)+size1;
locs3(:,3,:,:) = locs3(:,2,:,:)+size1;
locs3 = permute(locs3,[1 2 4 3]);
locs3 = vec(locs3);
method2 = matrix_withdraw(:,locs3);
method2 = reshape(method2,[88,9,64]);
method2 = permute(method2,[1 3 2]);

Method1和method2等效，并且呈现完全相同的结果。 Method2也比method1快10倍。我的问题是，有没有更清洁的方法？

Answer 1

如果拥有图像处理工具箱，则可以使用im2col将每个块作为一列，然后将reshape作为4-D数组：

blockSize = [3 3];
locs = reshape(im2col(matrix, blockSize), [blockSize size(matrix)-blockSize+1]);

或者您可以直接使用implicit singleton expansion构建一个4D索引数组。这不需要任何工具箱：

m = 3; n = 3; % block size
[M, N] = size(matrix);
ind = (1:m).'+(0:n-1)*M + reshape(0:M-m, 1, 1, []) + reshape((0:N-n)*M, 1, 1, 1, []);
locs = matrix(ind);

获取由每个子正方形组成的矩阵

1 个答案: