Matlab的：找到在2D最小像素具有特定的RGB

Question

讓假設我有此圖像：

我需要得到與RGB =（0，80，41）的最小像素。找到我會做的像素：

bart = imread('bart.jpg'); 

% Find pixels that have red=0, g = 80 or b = 41 
r_find = find(bart(:, :, 1) == 0); 
g_find = find(bart(:, :, 2) == 80); 
b_find = find(bart(:, :, 3) == 41); 

% intersect pixels to find pixels that his color are (0,80,41) 
find_res = intersect(intersect(r_find, g_find), b_find); 

%find the minimum pixel (1D) 
pixel = min(find_res); 

是的，我已經創建它，但我怎樣才能獲得像素（2D）的x，y座標？

Answer 1

您的錯誤是對每個顏色通道單獨使用find操作。

的解決方案是簡單的第一施加條件：

[row, col] = find(((bart(:, :, 1) == 0) & (bart(:, :, 2) == 80) & (bart(:, :, 3) == 41)), 1) 

上面的例子，最小化行第一座標。

如果你需要以最小化柱拳頭，可以施加find前轉置：

[col, row] = find([((bart(:, :, 1) == 0) & (bart(:, :, 2) == 80) & (bart(:, :, 3) == 41))]', 1) 

---

通過舉例說明：

%Read input image 
RGB = imread('https://i.stack.imgur.com/2sRiY.jpg'); 

%Unpack RGB planes (just for example purpose - you don't need to do it). 
R = RGB(:, :, 1); 
G = RGB(:, :, 2); 
B = RGB(:, :, 3); 

%(R == 0) is a logical matrix with 1 where condition is true, and 0 where false. 
%Same think for (G == 80) and for (B == 41) 
figure;imshow(R == 0); %Same as imshow(RGB(:,:,1) == 0) 
figure;imshow(G == 80); 
figure;imshow(B == 41); 

圖片：
ř== 0

摹== 80

乙== 41

%Now use AND operation between the three logical matrices: 
A = ((RGB(:, :, 1) == 0) & (RGB(:, :, 2) == 80) & (RGB(:, :, 3) == 41)); 

%A is a logical matrix with 1 where condition is true, and 0 where false. 
figure;imshow(A); 

圖像A：

%The following operation minimize column first: 
%Find first index where A equals true (where value equals 1). 
[col, row] = find(A, 1); 

%In case you need to minimize the row first, you can transpose A: 
[row, col] = find(A', 1); 

%All operations in single statement: 
[row, col] = find(((RGB(:, :, 1) == 0) & (RGB(:, :, 2) == 80) & (RGB(:, :, 3) == 41)), 1); 

Answer 2

您可以繪製圖像的對角線（使用poly2mask函數）並找到沿指定值對角線的第一個像素。

[rs ,cs,~] = size(bart); 
%create a mask image that has diagonal pixels value of 1 
mask = poly2mask([0 cs cs], [0 rs rs],rs, cs); 
%setting last argument to 1 to find the first pixel that meets the condition 
[r c] = find(mask & bart(:, :, 1) == 0 & bart(:, :, 2) == 80 & bart(:, :, 3) == 41,1) 

使用兩個角點找到對角線方程的效率更高，並且只能使用線上的像素。