Question

我有一个3d矩阵(3x3x3)，我需要提取3d补丁(2x2x2)并将其转换为矢量。

在2D中，简单地说：

I=randi(5,3,3);

2d_patches=im2col(I,[2 2],'sliding');

3D怎么样？

I=randi(5,3,3,3);

3d_patches= ???

im2col只适用于2d。在3d中，我应重新组合向量1和7，2和8，... 这项任务有快速功能吗？

Answer 1

我不相信有任何内置方法可以做到这一点。如果您需要快速，那么在c中编写自己的mex函数并从Matlab调用它应该相当简单。

这是我的（快速和肮脏）解决方案：

im3col.c：

#include <mex.h>

void mexFunction( int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[] )
{
  const mxArray *I = prhs[0];
  double *indata = mxGetPr(I);
  double *patchSize = mxGetPr(prhs[1]);
  const int *size = mxGetDimensions(I);
  int J = (int)patchSize[0], K = (int)patchSize[1], H = (int)patchSize[2];
  int M = size[0],           N = size[1],           P = size[2];

  int numPatches = (M - J + 1)*(N - K + 1)*(P - H + 1);
  int out_rows = J*K*H, out_cols = numPatches;
  mxArray *out = mxCreateDoubleMatrix( out_rows, out_cols, mxREAL );
  double *outdata = mxGetPr(out);

  int patch = 0;
  for( int h_offset = 0; h_offset < P-H+1; h_offset++ ){
    for( int k_offset = 0; k_offset < N-K+1; k_offset++ ){
      for( int j_offset = 0; j_offset < M-J+1; j_offset++ ){
        int row = 0;
        for( int h = 0; h < H; h++ ){
          for( int k = 0; k < K; k++ ){
            for( int j = 0; j < J; j++ ){
              outdata[patch*out_rows + row] = 
                indata[ (j_offset+j) + (k_offset+k)*M + (h_offset+h)*M*N ];
              ++row;
            }}}
      ++patch;
      }}}
  plhs[0] = out;
}

编译：

>> mex -O CFLAGS="\$CFLAGS -std=c99 -Wall" im3col.c

测试：

>> A(:,:,1) = [1,4,7;2,5,8;3,6,9]; A(:,:,2) = [10,13,16;11,14,17;12,15,18];
>> B = im3col(A, [2,2,1])

B =

     1     2     4     5    10    11    13    14
     2     3     5     6    11    12    14    15
     4     5     7     8    13    14    16    17
     5     6     8     9    14    15    17    18

>> A(:,:,1),A(:,:,2)

ans =

     1     4     7
     2     5     8
     3     6     9


ans =

    10    13    16
    11    14    17
    12    15    18

Answer 2

这是另一个方向：（它非常缓慢，肯定有更快的方式）

function [img] = patch2im_2d_time(patch, size_img, size_patch, size_skip, border)
Nx = size_img(1);
Ny = size_img(2);
Nt = size_img(5);

psz1 = size_patch(1);
psz2 = size_patch(2);
psz3 = size_patch(3);

%Extract blocks. One could save a lot here.
patches  = reshape(patch, [psz1 psz2 psz3 size(patch,2)]);
c = 1;
img2 = zeros(squeeze(size_img));
%Count for each pixel how many times we added smth to it.
add_count = zeros(size_img);

%The first three loops, loop through all the pixels in the image
for d=1:Nt-psz3+1
    for j=1:Nx-psz2+1
        for i=1:Ny-psz1+1
            %Here we get the next patch. The next patch is always
            %the patch that has the pixel at i,j,d at its top front corner.
            current_patch = patches(:,:,:,c);
            %counter for the next patch
            c = c + 1;
            %In this loop we add the patch values of each pixel in the
            %patch to the image. i,j,d is the base. We add the offset
            %ii jj and dd to it. This iteration takes psz^3 many
            %iterations.
            for dd=1:psz3
                for ii=1:psz2
                    for jj=1:psz1
                        img2(i+ii-1,j+jj-1,d+dd-1) = img2(i+ii-1,j+jj-1,d+dd-1) + current_patch(ii,jj,dd);
                        add_count(i+ii-1,j+jj-1,d+dd-1) = add_count(i+ii-1,j+jj-1,d+dd-1) + 1;
                    end
                end
            end

        end
    end
end
    img = flipud(rot90(img2 ./ add_count,1));
end

请记住，MATLAB使用col major。

Answer 3

%One possible way to use matlab to call im2col and reshape twice
%N = [row, col, num_frames]

[x_height, ~, num_frames] = size(N);
patchSize = 16;
patchTemporal = 10;

N = reshape(N, x_height, []);

N = im2col(N, [patchSize, patchSize], 'distinct');

N = reshape(N, [], num_frames);

N = im2col(N, [patchSize^2, patchTemporal], 'distinct');
% N = [patchSize^2 *patchTemporal x numPatches]

Answer 4

大家好，这个解决方案怎么样？从我建议获得3x3x3 ROI：

blkSize=3; % should be a odd value like 3,5,7,etc
r=floor(blkSize/2);
k=1;

for sliceNo=(r+1):(size(I,3)-r)

   img= I(:,:,sliceNo-r:sliceNo+r);

   noPix=(size(img,1)-2*r)*(size(img,2)-2*r);
   neiblk=zeros(blkSize^3,noPix);

   for blk=1:blkSize
       neiblk(blkSize^2*(blk-1)+1:blkSize^2*blk,:)=im2col(img(:,:,blk),...
             [blkSize,blkSize],'sliding');            
   end

   ROIs(:,noPix*(k-1)+1:noPix*k)=neiblk;
   k=k+1;
end

从3d矩阵的3d补丁

4 个答案:

im3col.c：

编译：

测试：