在matlab中的循环中平均矩阵的一个子集

Question

我使用我认为是矩阵的图像。

我想将 800 x 800 矩阵 (A) 转换为 400 x 400 矩阵 (B)，其中 A 矩阵的 4 个单元格的平均值 = B 矩阵的 1 个单元格（我知道这不正确代码行）：

B[1,1] =mean2(A[1,1 + 1,2 + 2,1 + 2,2])

整个矩阵依此类推...

B [1,2]=mean2(A[1,3 + 1,4 + 2,3 + 2,4 ])

我想：

1) 将 A 矩阵重塑为 2 x 320 000 矩阵，这样我就得到了我需要相邻平均的四个单元格，之后处理行号更容易。

Im4bis=reshape(permute(reshape(Im4,size(Im4,2),2,[]),[2,3,1]),2,[]);

2) 用我需要平均（子集化）的 4 个单元格创建一个单元格数组并计算它的平均值。这就是它不起作用的地方

I{1,160000}=ones,
for k=drange(1:2:319999)
    for n=1:160000
        I{n}=mean2(Im4bis(1:2,k:k+1));
    end
end

我创建了一个 400 x 400 个单元格的空矩阵（实际上是一个 1 x 160000 的向量），我想用平均值填充它，但我得到一个 1 x 319 999 个单元格的矩阵，其中一个单元格是 2 个单元格空.

寻找光线

我的输入图像：

Answer 1

让 A 表示您的矩阵，

m = 2; %// block size: rows
n = 2; %// block size: columns

方法一

使用blockproc:

B = blockproc(A, [m n], @(x) mean(x.data(:)));

示例：

>> A = magic(6)
A =
    35     1     6    26    19    24
     3    32     7    21    23    25
    31     9     2    22    27    20
     8    28    33    17    10    15
    30     5    34    12    14    16
     4    36    29    13    18    11
>> B = blockproc(A, [m n], @(x) mean(x.data(:)))
B =
   17.7500   15.0000   22.7500
   19.0000   18.5000   18.0000
   18.7500   22.0000   14.7500

方法二

如果您更喜欢重塑方式（这可能更快），请使用 this great answer 将矩阵组织成沿第三维平铺的 2x2 块，沿前两个维度取平均值，然后重塑结果：

T = permute(reshape(permute(reshape(A, size(A, 1), n, []), [2 1 3]), n, m, []), [2 1 3]);
B = reshape(mean(mean(T,1),2), size(A,1)/m, size(A,2)/n);

方法三

应用二维卷积（conv2) and then downsample. The convolution computes more entries than are really necessary (hence the downsampling), but on the other hand it can be done separably，这有助于加快速度：

B = conv2(ones(m,1)/m, ones(1,n)/n ,A,'same');
B = B(m-1:m:end ,n-1:n:end);

Answer 2

方法一

使用mat2cell and cellfun

AC = mat2cell(A, repmat(2,size(A,1)/2,1), repmat(2,size(A,2)/2,1));

out = cellfun(@(x) mean(x(:)), AC);

方法二

使用im2col

out = reshape(mean(im2col(A,[2 2],'distinct')),size(A)./2);

方法三

使用简单for循环

out(size(A,1)/2,size(A,2)/2) = 0;
k = 1;
for i = 1:2:size(A,1)
    l = 1;
    for j = 1:2:size(A,2)
        out(k,l) = mean(mean(A(i:i+1,j:j+1)));
        l = l+1;
    end
    k = k+1;
end

输入图像测试：

A = rgb2gray(imread('inputImage.png'));

%// Here, You could use any of the method from any answers 
%// or you could use the best method from the bench-marking tests done by Divakar
out = reshape(mean(im2col(A,[2 2],'distinct')),size(A)./2);  

imshow(uint8(out));  

imwrite(uint8(out),'outputImage.bmp');

输出图像：

通过读取已写入的图像进行最终检查

B = imread('outputImage.bmp');

>> whos B

Name        Size              Bytes  Class    Attributes

B         400x400            160000  uint8

Answer 3

基于 this solution 的一种方法，使用 reshape、sum 和 squeeze -

sublen = 2; %// subset length
part1 = reshape(sum(reshape(A,sublen,[])),size(A,1)/sublen,sublen,[]);
out = squeeze(sum(part1,2))/sublen^2;

基准测试

设置#1

以下是目前所列方法针对 800x 800 -

输入数据大小的运行时比较

%// Input
A = rand(800,800);

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

disp('----------------------- With  RESHAPE + SUM + SQUEEZE')
tic
sublen = 2; %// subset length
part1 = reshape(sum(reshape(A,sublen,[])),size(A,1)/sublen,sublen,[]);
out = squeeze(sum(part1,2))/sublen^2;
toc, clear sublen part1 out

disp('----------------------- With  BLOCKPROC')
tic
B = blockproc(A, [2 2], @(x) mean(x.data(:))); %// [m n]
toc, clear B

disp('----------------------- With  PERMUTE + MEAN + RESHAPE')
tic
m = 2;n = 2;
T = permute(reshape(permute(reshape(A, size(A, 1), n, []),...
      [2 1 3]), n, m, []), [2 1 3]);
B = reshape(mean(mean(T,1),2), size(A,1)/m, size(A,2)/m);
toc, clear B T m n

disp('----------------------- With  CONVOLUTION')
tic
m = 2;n = 2;
B = conv2(ones(m,1)/m, ones(1,n)/n ,A,'same');
B = B(m-1:m:end ,n-1:n:end);
toc, clear m n B

disp('----------------------- With  MAT2CELL')
tic
AC = mat2cell(A, repmat(2,size(A,1)/2,1), repmat(2,size(A,2)/2,1));
out = cellfun(@(x) mean(x(:)), AC);
toc

disp('----------------------- With  IM2COL')
tic
out = reshape(mean(im2col(A,[2 2],'distinct')),size(A)./2);
toc

运行时结果 -

----------------------- With  RESHAPE + SUM + SQUEEZE
Elapsed time is 0.004702 seconds.
----------------------- With  BLOCKPROC
Elapsed time is 6.039851 seconds.
----------------------- With  PERMUTE + MEAN + RESHAPE
Elapsed time is 0.006015 seconds.
----------------------- With  CONVOLUTION
Elapsed time is 0.002174 seconds.
----------------------- With  MAT2CELL
Elapsed time is 2.362291 seconds.
----------------------- With  IM2COL
Elapsed time is 0.239218 seconds.

为了使运行时间更公平，我们可以在最快的三种方法之上使用 1000 的多次试验，以获得相同的输入数据大小 800 x 800，从而得到 -

----------------------- With  RESHAPE + SUM + SQUEEZE
Elapsed time is 1.264722 seconds.
----------------------- With  PERMUTE + MEAN + RESHAPE
Elapsed time is 3.986038 seconds.
----------------------- With  CONVOLUTION
Elapsed time is 1.992030 seconds.

设置#2

以下是针对 10000x 10000 的较大输入数据大小的三种最快方法的运行时比较 -

----------------------- With  RESHAPE + SUM + SQUEEZE
Elapsed time is 0.158483 seconds.
----------------------- With  PERMUTE + MEAN + RESHAPE
Elapsed time is 0.589322 seconds.
----------------------- With  CONVOLUTION
Elapsed time is 0.307836 seconds.

在matlab中的循环中平均矩阵的一个子集

Average a subset of a matrix in a loop in matlab

matlab

image-processing

matrix

mean

subset

方法一

方法二

方法三

基准测试