Question

我有一个60 x 21 x 700矩阵，其中60 x 21代表pressure output x number of frames。我想找到生成每帧最大平均压力的2 x 2窗口，并将其存储在一个新变量中，以便绘制它。例如，如果矩阵看起来像这样：

01 02 02 01 01
02 01 01 02 02
02 03 04 04 03
01 02 06 10 05
02 02 08 09 05

2 x 2窗口的最大窗口及其平均值为 -

到目前为止，在我寻找解决方案时，我只能找到一种获取最大值的方法（例如，来自上面矩阵的10），但实际上无法工作如何从一个没有固定索引来引用的小区域获得最大平均值。我对MATLAB很新，所以如果我错过了某些内容或者根本没有正确理解事情，我表示歉意。任何帮助或指导将不胜感激。

Answer 1

2D阵列：对于给定的2D阵列输入，您可以使用2D convolution -

%// Perform 2D convolution with a kernel of `2 x 2` size with all ones
conv2_out = conv2(A,ones(2,2),'same')

%// Find starting row-col indices of the window that has the maximum conv value
[~,idx] = max(conv2_out(:))
[R,C] = ind2sub(size(A),idx)

%// Get the window with max convolution value
max_window = A(R:R+1,C:C+1)

%// Get the average of the max window
out =  mean2(max_window)

逐步运行代码 -

A =
     1     2     2     1     1
     2     1     1     2     2
     2     3     4     4     3
     1     2     6    10     5
     2     2     8     9     5
conv2_out =
     6     6     6     6     3
     8     9    11    11     5
     8    15    24    22     8
     7    18    33    29    10
     4    10    17    14     5
idx =
    14
R =
     4
C =
     3
max_window =
     6    10
     8     9
out =
         8.25

多维数组：对于多维数组，需要执行ND convolution -

%// Perform ND convolution with a kernel of 2 x 2 size with all ONES
conv_out = convn(A,ones(2,2),'same')

%// Get the average for all max windows in all frames/slices  
[~,idx] = max(reshape(conv_out,[],size(conv_out,3)),[],1)
max_avg_vals = conv_out([0:size(A,3)-1]*numel(A(:,:,1)) + idx)/4

%// If needed, get the max windows across all dim3 slices/frames
nrows = size(A,1)
start_idx = [0:size(A,3)-1]*numel(A(:,:,1)) + idx
all_idx = bsxfun(@plus,permute(start_idx(:),[3 2 1]),[0 nrows;1 nrows+1])
max_window = A(all_idx)

示例输入，输出 -

>> A
A(:,:,1) =
     4     1     9     9
     3     7     5     5
     9     6     1     6
     7     1     1     5
     4     2     2     1
A(:,:,2) =
     9     4     2     2
     3     6     4     5
     3     9     1     1
     6     6     8     8
     5     3     6     4
A(:,:,3) =
     5     5     7     7
     6     1     9     9
     7     7     5     4
     4     1     3     7
     1     9     3     1
>> max_window
max_window(:,:,1) =
     9     9
     5     5
max_window(:,:,2) =
     8     8
     6     4
max_window(:,:,3) =
     7     7
     9     9
>> max_avg_vals
max_avg_vals =
            7          6.5            8

Answer 2

移动平均线可以通过简单的卷积来完成。在你的情况下它必须是2D，所以：

A = [01 02 02 01 01
     02 01 01 02 02
     02 03 04 04 03
     01 02 06 10 05
     02 02 08 09 05];

B = [1 1;1 1] / 4 ; %// prepare moving average filter [2x2]

C = conv2(A,B,'valid') ; %// perform 2D moving average

产地：

C =
    1.5     1.5     1.5     1.5
    2       2.25    2.75    2.75
    2       3.75    6       5.5
    1.75    4.5     8.25    7.25

这正是每个[2x2]区域的平均值。

Answer 3

使用im2col

的另一种方法

%// getting each 2x2 sliding submatrices as columns
cols = im2col(A,[2 2],'sliding'); 

%// getting the mean & reshaping it to 2D matrix
C = reshape(mean(cols),size(A,1)-1,[]); 

%// to find the maximum of the sub-matrix-means and its corresponding index.
[B,i] = max(mean(cols)); 

%// reshaping the corresponding sub-matrix to 2D matrix
mat = reshape(cols(:,i),2,2);

结果：

>> mat

mat =

 6    10
 8     9

>> C

C =

1.5000    1.5000    1.5000    1.5000
2.0000    2.2500    2.7500    2.7500
2.0000    3.7500    6.0000    5.5000
1.7500    4.5000    8.2500    7.2500

>> B

B =

8.2500

滑动最大窗口及其多维数组的平均值

3 个答案: