opencv2.4.13中的最小值，最大值，平均值過濾器

Question

是否有任何內置函數將min，max和Avg過濾器應用於opencv2.4.13中的圖像？ 我正在使用C++。

Answer 1

由於@Miki在評論中提到，boxfilter是一個平均值過濾器。只需設置你想要的內核大小，並保留normalize=true（默認值）。

功能erode和dilate分別是最小和最大過濾器。您可以使用​​創建內核，創建您自己的內核，或像我一樣使用默認的3x3。對於erode和-inf，對於dilate，邊界默認設置爲+inf，因此它們對結果沒有貢獻。

int main(int argc, const char * argv[]) { 

    char image_data[25] = {1, 3, 8, 8, 4, 
          4, 2, 7, 9, 9, 
          1, 5, 0, 5, 9, 
          3, 7, 5, 2, 1, 
          0, 4, 7, 9, 4}; 
    cv::Mat image = cv::Mat(5, 5, CV_8U, image_data); 
    std::cout << "image = " << std::endl << image << std::endl; 

    cv::Mat avgImage; 
    // Perform mean filtering on image using boxfilter 
    cv::boxFilter(image, avgImage, -1, cv::Size(3,3)); 
    std::cout << "avgImage = " << std::endl << avgImage << std::endl; 

    cv::Mat kernel; // Use the default structuring element (kernel) for erode and dilate 

    cv::Mat minImage; 
    // Perform min filtering on image using erode 
    cv::erode(image, minImage, kernel); 
    std::cout << "minImage = " << std::endl << minImage << std::endl; 

    cv::Mat maxImage; 
    // Perform max filtering on image using dilate 
    cv::dilate(image, maxImage, kernel); 
    std::cout << "maxImage = " << std::endl << maxImage << std::endl; 

    return 0; 
} 

下面是結果：

image = 
[ 1, 3, 8, 8, 4; 
    4, 2, 7, 9, 9; 
    1, 5, 0, 5, 9; 
    3, 7, 5, 2, 1; 
    0, 4, 7, 9, 4] 
avgImage = 
[ 3, 4, 6, 8, 8; 
    3, 3, 5, 7, 7; 
    4, 4, 5, 5, 6; 
    4, 4, 5, 5, 5; 
    5, 5, 5, 4, 4] 
minImage = 
[ 1, 1, 2, 4, 4; 
    1, 0, 0, 0, 4; 
    1, 0, 0, 0, 1; 
    0, 0, 0, 0, 1; 
    0, 0, 2, 1, 1] 
maxImage = 
[ 4, 8, 9, 9, 9; 
    5, 8, 9, 9, 9; 
    7, 7, 9, 9, 9; 
    7, 7, 9, 9, 9; 
    7, 7, 9, 9, 9]