我想实现一个算法，需要一个圆形的滑动窗口，即经过图像的所有像素，并为每个窗口，我需要提取躺在直径的圆的不同角度只能像素。

I'm trying to implement an algorithm that needs a circular sliding window, that goes through all the pixels of the image, and for each window I need to extract only pixels that lay on diameters at different angles of the circle.

我试图解释与使用上图中的更好。试想一下，这是一个圆形的滑动窗口的结果，我想只有躺在红线（直径倾斜的PI / 8）的像素值。

I try to explain better with the use of the image above. Imagine that this is the result of a circular sliding window, I want to get only the pixel values that lay on the red line (diameter tilted of pi/8).

到目前为止，我写了这些线路的code：

So far, I wrote these lines of code:

I= imread('lena.png'); 
I = im2double(I);
h = fspecial('disk',5);
h = h > 0;
dir = pi/8;
o_image = blockproc(I, [1 1], @(x) MMF2D(x,h,dir), 'BorderSize', [5 5],...
'TrimBorder', false, 'PadPartialBlocks', true);

和功能 MMF2D ：

function [ o_pixel ] = MMF2D( i_block, i_window, i_directions)
%MMF2D Summary of this function goes here
%   Detailed explanation goes here

new_block = i_block.data .* i_window;

但在这里，我不知道如何继续获取像素躺在直径。直径可倾斜任意角度的。任何帮助是极大AP preciated！

But from here, I don't know how to continue for getting pixels that lay on diameter. The diameter can be tilted of any angle. Any help is greatly appreciated!

推荐答案

这就是我想要做的：

首先创建蒙（你可以把这个变成一个功能，如果你想），并获得相关的像素指标为角度的功能：

First create the mask (you can put this into a function if you want), and get the relevant pixels indices as function of angle:

%% create mask and get indices as function of angle
o=5;
m=zeros(2*o+1);
m(o+1,:)=1;
theta=0:15:90; % in degrees, theres no need to go beyond 90 deg becuase of symmetry
for n=1:numel(theta)
   id{n}=find(imrotate(m,theta(n),'nearest','crop'));
end;

我用一个单元阵列，因为可以有不同数量的每个角指标。

I use a cell array because there can be different number of indices per angle.

然后读取图片

I= imread('https://m.xsw88.com/allimgs/daicuo/20230911/3061.png'); 

重新排列图像块成列

Rearrange image blocks into columns

B = im2col(I,[2*o+1 2*o+1],'sliding');

你想要的仅仅是：

what you want is just:

for n=1:numel(theta)
    C{n} =  B(id{n},:);
end

在C中重新$ P $每个电池单元psents这是一个直径长度 2 * O + 1 以一定的角度 THETA像素这是每定义的，用于图像中的每个像素。 因此， 13 C {1} 会给你所有的像素 THETA = 0 和Ç {2} 为 THETA = 15 等等...

each cell element in C represents the pixels that were on a diameter of length 2*o+1 at an angle theta that was per-defined, for each pixel in the image. So C{1} will give you all the pixels for theta=0, and C{2} for theta=15 etc...