c++ - 如何选择阈值自动使用直方图的峰值？

Question

通过 OpenCV 库，我想对这样的图像进行阈值处理:

threshold(image, thresh, 220, 255, THRESH_BINARY_INV)

但是我想自动找到阈值(220)。

我使用 Otsu 来估计阈值。但它在我的情况下不起作用。

因此，我应该使用直方图峰值技术。我想在直方图中找到对应于图像背景和对象的两个峰值。它会自动设置两个峰值中间的阈值。我使用这本书(第 117 页和 496-505 页):Dwayne Phillips 的“C 图像处理”(http://homepages.inf.ed.ac.uk/rbf/BOOKS/PHILLIPS/)。我使用源代码在直方图中找到对应于图像背景和对象的两个峰值。这是我的形象:

这是我的 C++ 代码:

#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/opencv.hpp>

#include <iostream>
#include <stdio.h>
#include <fstream>

using namespace std;
using namespace cv;


int main()
{
    Mat image0 = imread("C:/Users/Alireza/Desktop/contrast950318/2.bmp");
    imshow("image0", image0);

    Mat image, thresh, Tafrigh;
    cvtColor(image0, image, CV_RGB2GRAY);


    int N = image.rows*image.cols;

    int histogram[256];

    for (int i = 0; i < 256; i++) {
        histogram[i] = 0;
    }

    //create histo
    for (int i = 0; i < image.rows; i++){
        for (int j = 0; j < image.cols; j++){
            histogram[((int)image.at<uchar>(i, j))]++;
        }
    }


    int peak1, peak2;
    #define PEAKS 30
    int distance[PEAKS], peaks[PEAKS][2];

    int i, j = 0, max = 0, max_place = 0;

    for (int i = 0; i<PEAKS; i++){
        distance[i] = 0;
        peaks[i][0] = -1;
        peaks[i][1] = -1;
    }

    for (i = 0; i <= 255; i++){
        max = histogram[i];
        max_place = i;

        //insert_into_peaks(peaks, max, max_place);
        //int max, max_place, peaks[PEAKS][2];
        //int i, j;
        /* first case */
        if (max > peaks[0][0]){
            for (i = PEAKS - 1; i > 0; i--){
                peaks[i][0] = peaks[i - 1][0];
                peaks[i][1] = peaks[i - 1][1];
            }
            peaks[0][0] = max;
            peaks[0][1] = max_place;
        } /* ends if */

        /* middle cases */
        for (j = 0; j < PEAKS - 3; j++){
            if (max < peaks[j][0] && max > peaks[j + 1][0]){
                for (i = PEAKS - 1; i > j + 1; i--){
                    peaks[i][0] = peaks[i - 1][0];
                    peaks[i][1] = peaks[i - 1][1];
                }

                peaks[j + 1][0] = max;
                peaks[j + 1][1] = max_place;
            } /* ends if */
        } /* ends loop over j */
        /* last case */
        if (max < peaks[PEAKS - 2][0] &&
            max > peaks[PEAKS - 1][0]){
            peaks[PEAKS - 1][0] = max;
            peaks[PEAKS - 1][1] = max_place;
        } /* ends if */

}/* ends loop over i */



    for (int i = 1; i<PEAKS; i++){
        distance[i] = peaks[0][1] - peaks[i][1];
        if (distance[i] < 0)
            distance[i] = distance[i] * (-1);
    }

    peak1 = peaks[0][1];
    cout << "  peak1= " << peak1;

    for (int i = PEAKS - 1; i > 0; i--){
        if (distance[i] > 1)
            peak2 = peaks[i][1];

    }
    cout << "  peak2= " << peak2;


    int mid_point;
    //int peak1, peak2;
    short hi, low;

    unsigned long sum1 = 0, sum2 = 0;
    if (peak1 > peak2)
        mid_point = ((peak1 - peak2) / 2) + peak2;
    if (peak1 < peak2)
        mid_point = ((peak2 - peak1) / 2) + peak1;

    for (int i = 0; i<mid_point; i++)
        sum1 = sum1 + histogram[i];
    for (int i = mid_point; i <= 255; i++)
        sum2 = sum2 + histogram[i];
    if (sum1 >= sum2){
        low = mid_point;
        hi = 255;
    }
    else{
        low = 0;
        hi = mid_point;
    }

    cout << "  low= " << low << "  hi= " << hi;



double  threshnum = 0.5* (low  + hi);
threshold(image, thresh, threshnum, hi, THRESH_BINARY_INV);


    waitKey(0);
    return 0;
}

但我不知道这段代码是否正确。如果正确，为什么阈值是202？

对于如何解决这个任务，您有什么建议？或者我可以在互联网上的哪些资源上找到帮助？

Answer 1

您也可以使用最大熵。在某些情况下，仅使用熵的高频可能会更好

int maxentropie(const cv::Mat1b& src)
{
    // Histogram
    cv::Mat1d hist(1, 256, 0.0);
    for (int r=0; r<src.rows; ++r)
        for (int c=0; c<src.cols; ++c)
            hist(src(r,c))++;

    // Normalize
    hist /= double(src.rows * src.cols);

    // Cumulative histogram
    cv::Mat1d cumhist(1, 256, 0.0);
    float sum = 0;
    for (int i = 0; i < 256; ++i)
    {
        sum += hist(i);
        cumhist(i) = sum;
    }

    cv::Mat1d hl(1, 256, 0.0);
    cv::Mat1d hh(1, 256, 0.0);

    for (int t = 0; t < 256; ++t)
    {
        // low range entropy
        double cl = cumhist(t);
        if (cl > 0)
        {
            for (int i = 0; i <= t; ++i)
            {
                if (hist(i) > 0)
                {
                    hl(t) = hl(t) - (hist(i) / cl) * log(hist(i) / cl);
                }
            }
        }

        // high range entropy
        double ch = 1.0 - cl;  // constraint cl + ch = 1
        if (ch > 0)
        {
            for (int i = t+1; i < 256; ++i)
            {
                if (hist(i) > 0)
                {
                    hh(t) = hh(t) - (hist(i) / ch) * log(hist(i) / ch);
                }
            }
        }
    }

    // choose best threshold

    cv::Mat1d entropie(1, 256, 0.0);
    double h_max = hl(0) + hh(0);
    int threshold = 0;
    entropie(0) = h_max;

    for (int t = 1; t < 256; ++t)
    {
        entropie(t) = hl(t) + hh(t);
        if (entropie(t) > h_max)
        {
            h_max = entropie(t);
            threshold = uchar(t);
        }
    }
    if(threshold==0) threshold=255;
    return threshold;
}