矩阵的滑动窗口求和 [英] Sliding window summation of a matrix

问题描述

我有一个50x50的矩阵，我想对每10x10(或另一个设置大小的值-始终为正方形)重叠网格中的值求和，即:

I have a 50x50 matrix, and I'd like to sum up the values in every 10x10 (or another set size value - always square) overlapping grid i.e.:

为清楚起见，重叠窗口仅以对角线显示.我尝试做的第一个任务是定义每个窗口的坐标:

Overlapping windows are shown only in the diagonal for the sake of clarity. The first task I've tried to do is define the coordinates of each window:

win=10;
start = [1,10,1,10];
for y=1:(50-win)
    for g=1:(50-win)
        tmp = [start(g,1)+1,start(g,2)+1,start(end,3),start(end,4)];
        start = [start;tmp];
    end
    start(end+1,1:4) = [1,10,1+y,10+y];
end

然后我将对每个窗口使用sum和逻辑索引遍历坐标列表.

And then I'd loop over the list of coordinates, using sum and logical indexing for each window.

问题1 :上面的代码并不是特别雄辩.任何人都可以展示出更"MATLABesque"的方式或更简洁的方式吗?

PROBLEM #1: The above code is not particularly eloquent. Can anybody show a more 'MATLABesque' way of doing it or a more concise way?

问题2 :然后，我想在矩阵中定义特定的坐标(索引)，例如m(26,26)并获取此坐标包含在其中的所有窗口的列表.但是我不知道该怎么做.有人可以告诉我怎么做吗?

PROBLEM #2: I'd then like to define a particular coordinate (index) in the matrix e.g. m(26,26) and get a list of all windows this coordinate is contained within. But I have no idea how to do this. Can anybody show me how?

推荐答案

问题1

我能想到的最类似于Matlab的方式是二维卷积(

Problem #1

The most Matlab-like way for doing this I can think of is two-dimensional convolution (conv2) (as I now see was commented by @rahnema1):

M = randi(9, 5, 5); % input: square matrix, arbitrary size
N = 3; % block size, assumed square, not larger than M
result = conv2(M, ones(N), 'valid');

等效地，您可以使用最近引入的 movsum 函数，两次(每个维度一次):

Equivalently, you can use the recently introduced movsum function, twice (once for each dimension):

result = movsum(movsum(M, N, 1, 'Endpoints', 'discard'), N, 2, 'Endpoints', 'discard');

示例:

M =
     4     4     3     1     2
     2     8     7     1     6
     3     6     7     5     5
     6     5     4     8     1
     5     9     6     9     4

result =
    44    42    37
    48    51    44
    51    59    49

问题2

最简单的方法(不是最有效的方法)是再次使用卷积，将逻辑矩阵包含在所需位置的true，否则包含false，并检查卷积不为零的地方:

Problem #2

The simplest way (not the most efficient one) is to use convolution again with a logical matrix containing true at the desired position and false otherwise, and checking where the convolution is not zero:

in_coords = [3 4]; % example input coordinates
T = false(size(M)); % initiallize matrix containing false, same size as M
T(in_coords(1), in_coords(2)) = true; % true at the desired coordinates
C = conv2(T, ones(N), 'valid'); % this gives 1 for blocks affected by in_coords
[ii, jj] = find(C); % row and column indices of nonzero values 
out_coords = [ii jj]; % build result

在此示例中，

out_coords =
     1     2
     2     2
     3     2
     1     3
     2     3
     3     3

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