我有一个 60 x 21 x 700
矩阵,其中 60 x 21
代表一个压力输出
x 帧数
。我想找到每帧生成最大平均压力的 2 x 2
窗口,并将其存储在一个新变量中,以便可以绘制它。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 -
06 10
08 09
= 8.25
到目前为止,在我寻找解决方案的过程中,我只能找到一种方法来获得最大值(例如,上面矩阵中的 10
),但真的无法解决如何从没有固定索引引用的小区域获得最大平均值。我是 MATLAB 的新手,所以如果我错过了什么或者只是没有正确理解事情,我深表歉意。
任何帮助或指导将不胜感激。
最佳答案
二维数组:对于给定的二维数组输入,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)
代码的逐步运行示例 -
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:对于多维数组的情况,你需要执行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
关于matlab - 多维数组的滑动最大窗口及其平均值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30072284/