cv :: Mat序列的像素中位数

Question

注意:我不是在询问Median Filter.

我有一系列图像让我们说:

std::array<cv::Mat,N> sequence;

我想将所有这些图像合二为一.这一张图片应满足:

新图像的每个像素是来自序列的其对应像素的中值.换一种说法:

Result(i,j)=median(sequence[0](i,j), sequence[1](i,j), ..., sequence[N](i,j));

这样做有内置功能吗？什么是最快的方式？

我到目前为止尝试过:迭代所有序列中的每个像素并排序然后取中位数然后将其存储在结果中.但是,这太过分了.

Answer 1

您可以使用直方图计算每个位置的顺序中位数.

假设您正在使用Mat1b图像,每个直方图将具有256个值.您需要存储直方图以及所有箱的总和:

struct Hist {
    vector<short> h;
    int count;
    Hist() : h(256, 0), count(0) {};
};

中值是直方图中对应于一半像素的索引count / 2.来自Rosetta代码的片段:

int i;
int n = hist.count / 2; // 'hist' is the Hist struct at a given pixel location
for (i = 0; i < 256 && ((n -= hist.h[i]) >= 0); ++i);
// 'i' is the median value

添加或删除图像时,更新每个像素位置的直方图,然后重新计算中值.此操作非常快,因为您不需要排序.

这有一些缺点:

1. 这仅适用于uchar值,否则每个直方图的长度将太大
2. 这种方法将使用大量的RAM,因为它需要rows x cols直方图.它可能不适用于大图像.
3. 以下实现适用于单通道图像,但可以轻松扩展到3个通道.

4. 您可以使用基于两个堆或近似方法的方法.你可以在这里找到细节:

   • 从整数流中查找运行中位数
   • C中的滚动中值算法
   • 从不断增长的集合中找出中值

码:

#include <vector>
#include <opencv2/opencv.hpp>

using namespace std;
using namespace cv;

struct Hist {
    vector<short> h;
    int count;
    Hist() : h(256, 0), count(0) {};
};

void addImage(vector<Mat1b>& images, Mat1b& img, vector<vector<Hist>>& M, Mat1b& med)
{
    assert(img.rows == med.rows);
    assert(img.cols == med.cols);

    for (int r = 0; r < img.rows; ++r) {
        for (int c = 0; c < img.cols; ++c){

            // Add pixel to histogram
            Hist& hist = M[r][c];
            ++hist.h[img(r, c)];
            ++hist.count;

            // Compute median
            int i;
            int n = hist.count / 2;
            for (i = 0; i < 256 && ((n -= hist.h[i]) >= 0); ++i);

            // 'i' is the median value
            med(r,c) = uchar(i);
        }
    }

    // Add image to my list
    images.push_back(img.clone());
}

void remImage(vector<Mat1b>& images, int idx, vector<vector<Hist>>& M, Mat1b& med)
{
    assert(idx >= 0 && idx < images.size());

    Mat1b& img = images[idx];
    for (int r = 0; r < img.rows; ++r) {
        for (int c = 0; c < img.cols; ++c){

            // Remove pixel from histogram
            Hist& hist = M[r][c];
            --hist.h[img(r, c)];
            --hist.count;

            // Compute median
            int i;
            int n = hist.count / 2;
            for (i = 0; i < 256 && ((n -= hist.h[i]) >= 0); ++i);

            // 'i' is the median value
            med(r, c) = uchar(i);
        }
    }

    // Remove image from list
    images.erase(images.begin() + idx);
}

void init(vector<vector<Hist>>& M, Mat1b& med, int rows, int cols)
{
    med = Mat1b(rows, cols, uchar(0));
    M.resize(rows);
    for (int i = 0; i < rows; ++i) {
        M[i].resize(cols);
    }
}


int main()
{
    // Your images... be sure that they have the same size
    Mat1b img0 = imread("path_to_image", IMREAD_GRAYSCALE);
    Mat1b img1 = imread("path_to_image", IMREAD_GRAYSCALE);
    Mat1b img2 = imread("path_to_image", IMREAD_GRAYSCALE);
    resize(img0, img0, Size(800, 600));
    resize(img1, img1, Size(800, 600));
    resize(img2, img2, Size(800, 600));



    int rows = img0.rows;
    int cols = img0.cols;

    vector<Mat1b> images;   // All your images, needed only if you need to remove an image
    vector<vector<Hist>> M; // histograms
    Mat1b med;              // median image

    // Init data strutctures
    init(M, med, rows, cols);

    // Add images. 'med' will be the median image and will be updated each time
    addImage(images, img0, M, med);
    addImage(images, img1, M, med);
    addImage(images, img2, M, med);

    // You can also remove an image from the median computation
    remImage(images, 2, M, med);

    // Hey, same median as img0 and img1 ;D

    return 0;
}