如何找到一個像素的鄰域和它的座標

Question

我需要找到像素的所有黑鄰域;對於我嘗試以下，但輸出不correct.Cropped輸入圖像是 標記圖像（我把顏色到該標籤） 我想添加的輸入圖像的黑色區域（區域周圍像橢圓結構）和改變周圍的橢圓形白色區域標記的圖像爲藍色（需要改變周圍的白色區域爲1）的標籤

clc 
    clear all;close all 
    img=[0 255 0 0 0; 
     0 1 1 255 0; 
     255 0 0 255 0;]; 
    [m n]=size(img) 
    img= padarray(img, [2 2], 0) 
    fi_neighbors=zeros(1,8); 
    for i=1:m 
     for j=1:n 
    for r=3:m 
     for c=3:n 

      neighbors(1) = img(r-1,c-1); % Upper left. 
    neighbors(2) = img(r-1,c); % Upper middle. 
    neighbors(3) = img(r-1,c+1); % Upper right. 
    neighbors(4) = img(r,c-1); % left. 
    neighbors(5) = img(r,c+1); % right. 
    neighbors(6) = img(r+1,c+1); % Lowerleft. 
    neighbors(7) = img(r+1,c); % lower middle. 
    neighbors(8) = img(r+1,c-1); % Lower left. 
     end 

    end 
    fi_neighbors=[neighbors;fi_neighbors] 
     end 
    end 

Answer 1

有在你的代碼的幾個錯誤，所以我會盡量採取他們一個接一個。這裏是你的原代碼，重新格式化了一下：

img = [0 0 0 0 0; 
     0 1 1 0 0; 
     0 0 0 0 0;]; 
[m n] = size(img) 
img = padarray(img, [2 2], 0) 
fi_neighbors = zeros(1,8); 

for i = 1:m 
    for j = 1:n 
     for r = 3:m 
     for c = 3:n 
      neighbors(1) = img(r-1,c-1); % Upper left. 
      neighbors(2) = img(r-1,c); % Upper middle. 
      neighbors(3) = img(r-1,c+1); % Upper right. 
      neighbors(4) = img(r,c-1); % left. 
      neighbors(5) = img(r,c+1); % right. 
      neighbors(6) = img(r+1,c+1); % Lower right. <-- corrected 
      neighbors(7) = img(r+1,c); % lower middle. 
      neighbors(8) = img(r+1,c-1); % Lower left. 
     end 
     end 
     fi_neighbors = [neighbors;fi_neighbors] 
    end 
end 

第一個錯誤是，循環變量i和j從不使用。因此它們是無用的（除了讓我們執行相同的計算m x n次）。讓我們把它們拿出來。

img = [0 0 0 0 0; 
     0 1 1 0 0; 
     0 0 0 0 0;]; 
[m n] = size(img) 
img = padarray(img, [2 2], 0) 
fi_neighbors = zeros(1,8); 

for r = 3:m 
    for c = 3:n 
     neighbors(1) = img(r-1,c-1); % Upper left. 
     neighbors(2) = img(r-1,c); % Upper middle. 
     neighbors(3) = img(r-1,c+1); % Upper right. 
     neighbors(4) = img(r,c-1); % left. 
     neighbors(5) = img(r,c+1); % right. 
     neighbors(6) = img(r+1,c+1); % Lower right. 
     neighbors(7) = img(r+1,c); % lower middle. 
     neighbors(8) = img(r+1,c-1); % Lower left.    
    end 
end 
fi_neighbors = [neighbors;fi_neighbors] 

隨着這些循環的方式進行，我們可以看到，你只更新fi_neighbors數組，你已經做了所有的計算後的一次。這就是爲什麼你之前得到相同結果m x n次。我們需要移動的循環，結果就是每個組計算後更新內部（我已經離開了這個設定的時間和僅列循環）：

for r = 3:m 
    for c = 3:n 
     neighbors(1) = img(r-1,c-1); % Upper left. 
     neighbors(2) = img(r-1,c); % Upper middle. 
     neighbors(3) = img(r-1,c+1); % Upper right. 
     neighbors(4) = img(r,c-1); % left. 
     neighbors(5) = img(r,c+1); % right. 
     neighbors(6) = img(r+1,c+1); % Lower right. 
     neighbors(7) = img(r+1,c); % lower middle. 
     neighbors(8) = img(r+1,c-1); % Lower left.    
     fi_neighbors = [neighbors;fi_neighbors] %// moved inside loops 
    end 
end 

接下來是我們的循環範圍。您在每一邊填充了2個零的數組，因此您從行/列3開始。這很好，但你仍然在m和n處結束循環。這意味着你處理(m-2) x (n-2)像素。你需要擴大範圍。但實際上，你的填充量超過了你的需要。每邊一個像素的填充就足夠了，所以讓我們一次都做，改變填充和糾正範圍：

img = [0 0 0 0 0; 
     0 1 1 0 0; 
     0 0 0 0 0;]; 
[m n] = size(img) 
img = padarray(img, [1 1], 0) 
fi_neighbors = zeros(1,8); 

for r = 2:m+1 
    for c = 2:n+1 
     neighbors(1) = img(r-1,c-1); % Upper left. 
     neighbors(2) = img(r-1,c); % Upper middle. 
     neighbors(3) = img(r-1,c+1); % Upper right. 
     neighbors(4) = img(r,c-1); % left. 
     neighbors(5) = img(r,c+1); % right. 
     neighbors(6) = img(r+1,c+1); % Lower right. 
     neighbors(7) = img(r+1,c); % lower middle. 
     neighbors(8) = img(r+1,c-1); % Lower left.    
     fi_neighbors = [neighbors;fi_neighbors] 
    end 
end 

現在你應該有一個工作計劃，讓你正確的輸出：

fi_neighbors = 

    0 0 0 0 0 0 0 0 <-- img(m,n) neighbors 
    1 0 0 0 0 0 0 0 
    1 1 0 0 0 0 0 0 
    0 1 1 0 0 0 0 0 
    0 0 1 0 0 0 0 0 
    0 0 0 0 0 0 0 0 
    0 0 0 1 0 0 0 0 
    0 0 0 1 0 0 0 0 
    0 0 0 0 1 0 0 0 
    0 0 0 0 1 0 0 0 
    0 0 0 0 0 0 0 0 
    0 0 0 0 0 0 0 1 
    0 0 0 0 0 0 1 1 
    0 0 0 0 0 1 1 0 
    0 0 0 0 0 1 0 0 <-- img(1,1) neighbors 
    0 0 0 0 0 0 0 0 

這可以工作，但看看我添加到輸出的筆記。結果的第一行包含像素img(m,n)的鄰居：圖像中的最後像素。倒數第二行包含img(1,1)的鄰居，最後一行由於您初始化爲zeros(1,8)而剛剛掛在那裏。所有這些對我來說都不自然，所以我要翻轉結果數組，以便img(1,1)是第一行。而且，在每個循環迭代中增長一個數組是很慢的。我們想要預分配數組，然後將數據放在適當的位置。

由於我們引用原始圖像中的像素，但是在填充圖像上計算，添加計數器變量比執行所有行/列索引計算容易得多，所以我添加了pix作爲計數器：

img = [0 0 0 0 0; 
     0 1 1 0 0; 
     0 0 0 0 0;]; 
[m n] = size(img); 
img = padarray(img, [1 1], 0) 
fi_neighbors = zeros(m*n,8); % preallocate result array 

pix = 0; % pixel counter 
for r = 2:m+1 
    for c = 2:n+1 
     pix = pix + 1; % find neighbors for next pixel 
     neighbors(1) = img(r-1,c-1); % Upper left. 
     neighbors(2) = img(r-1,c); % Upper middle. 
     neighbors(3) = img(r-1,c+1); % Upper right. 
     neighbors(4) = img(r,c-1); % left. 
     neighbors(5) = img(r,c+1); % right. 
     neighbors(6) = img(r+1,c+1); % Lower right. 
     neighbors(7) = img(r+1,c); % lower middle. 
     neighbors(8) = img(r+1,c-1); % Lower left. 
     fi_neighbors(pix,:) = neighbors; % update the correct row 
    end 
end 

結果：

fi_neighbors = 

    0 0 0 0 0 1 0 0 
    0 0 0 0 0 1 1 0 
    0 0 0 0 0 0 1 1 
    0 0 0 0 0 0 0 1 
    0 0 0 0 0 0 0 0 
    0 0 0 0 1 0 0 0 
    0 0 0 0 1 0 0 0 
    0 0 0 1 0 0 0 0 
    0 0 0 1 0 0 0 0 
    0 0 0 0 0 0 0 0 
    0 0 1 0 0 0 0 0 
    0 1 1 0 0 0 0 0 
    1 1 0 0 0 0 0 0 
    1 0 0 0 0 0 0 0 
    0 0 0 0 0 0 0 0 

現在，img(1,1)鄰居們都在第1行的img(m,n)鄰居們都行m * n個。

最後，您可以進行一些快速優化。他們都需要改變結果的順序。如果你願意以不同的順序呈現像素和鄰居，我將試圖清楚地表明，那麼我們可以首先改變循環的順序以利用MATLAB的列主要順序。

相反的：

for r = 2:m+1 
    for c = 2:n+1 
     ... 
    end 
end 

，我們將使用：

for c = 2:n+1 % be sure to swap m and n as well! 
    for r = 2:m+1 
     ... 
    end 
end 

對於大型矩陣，這應該與緩存保持我們正在當前列有幫助。這將改變像素的排序，以便代替第二像素是img(1,2)，現在將是img(2,1)，然後img(3,1)，直到img(m,1)，然後是img(1,2)。

第二個變化凝聚所有的鄰居計算爲2行（也許3），但它改變，因爲我將在下面顯示的順序：

pix = 0; % pixel counter 
for c = 2:n+1 
    for r = 2:m+1 
     pix = pix + 1; % find neighbors for next pixel 

     neighbors = img(r-1:r+1,c-1:c+1); 
     neighbors = neighbors(:).'; 

     fi_neighbors(pix,:) = neighbors; % update the correct row 
    end 
end 

這需要的img子矩陣當前像素周圍，然後使用冒號表示將其轉換爲列向量，最後將其轉換爲行向量。下面是導致鄰居位置的內neighbors：

neighbors(1) % Upper left. 
neighbors(2) % left. 
neighbors(3) % Lower left. 
neighbors(4) % Upper middle. 
neighbors(5) % *** current pixel *** 
neighbors(6) % Lower middle. 
neighbors(7) % Upper right. 
neighbors(8) % right. 
neighbors(9) % Lower right. 

但是現在我們已經在neighbors 9個元素包括當前像素！如果您確實不想列出當前像素，我們可以修復該問題，但無論哪種方式，我們都需要將fi_neighbors的大小更改爲(m*n,9)。如果需要，我們可以從結果中刪除第5列。

下面是最終代碼：

img = [0 0 0 0 0; 
     0 1 1 0 0; 
     0 0 0 0 0;]; 
[m n] = size(img); 
img = padarray(img, [1 1], 0); 
fi_neighbors = zeros(m*n,9); % preallocate result array 

pix = 0; % pixel counter 
for c = 2:n+1 
    for r = 2:m+1 
     pix = pix + 1; % find neighbors for next pixel 
     neighbors = img(r-1:r+1,c-1:c+1); 
     neighbors = neighbors(:).'; 
     fi_neighbors(pix,:) = neighbors; % update the correct row 
    end 
end 

fi_neighbors(:,5) = []; % OPTIONAL - remove current pixel from each row 

顯然，還有更多你可以做，但是這應該是一個很好的開始。