计算绘制图像灰度直方图

OpenCV中分析绘制直方图并没有直接可用的函数，以下是绘制直方图统计的一种实现。
paint_histogram.h：

#ifndef OPENVCV_H
#define  OPENVCV_H
#include <opencv2/opencv.hpp>
class Histogram1D
{
public:
    Histogram1D();
    ~Histogram1D();
    cv::Mat GetHistogram(const cv::Mat& image);
    cv::Mat GetHistogramImage(const cv::Mat& image, int zoom = 1);
    cv::Mat GetImageOfHistogram(const cv::Mat& hist, int zoom);
private:
    int histSize[1];
    float hranges[2];
    const float* ranges[1];
    int channels[1];
};
#endif

paint_histogram.cpp如下所示：

#include "paint_histogram.h"
using namespace cv;
using namespace std;

Histogram1D::Histogram1D()
{
    histSize[0] = 256;
    hranges[0] = 0.0;
    hranges[1] = 256.0;
    ranges[0] = hranges;
    channels[0] = 0;
}

Histogram1D::~Histogram1D()
{
}

Mat
Histogram1D::GetHistogram(const Mat& image)
{
    Mat hist;
    calcHist(&image, 1, channels, Mat(), hist, 1, histSize, ranges);
    return hist;
}

Mat
Histogram1D::GetHistogramImage(const cv::Mat& image, int zoom)
{
    Mat hist = GetHistogram(image);
    return GetImageOfHistogram(hist, zoom);
}
Mat
Histogram1D::GetImageOfHistogram(const cv::Mat& hist, int zoom)
{
    double maxVal = 0;
    double minVal = 0;

    minMaxLoc(hist, &minVal, &maxVal, 0, 0);

    int histSize = hist.rows;
    Mat histImg(histSize * zoom, histSize * zoom, CV_8U, Scalar(255));
    int hpt = static_cast<int>(0.9 * histSize);

    for (int h = 0; h < histSize; ++h) {
        float binVal = hist.at<float>(h);
        if (binVal > 0) {
            int intensity = static_cast<int>(binVal * hpt / maxVal);
            line(histImg, Point(h * zoom, histSize * zoom), Point(h * zoom, (histSize - intensity) * zoom), Scalar(0), zoom);
        }
    }
    return histImg;
}

main函数调用如下：

#include "paint_histogram.h"
using namespace std;
using namespace cv;

int main()
{

    Mat src = imread("flower.png", IMREAD_GRAYSCALE);
    Histogram1D h;
    namedWindow("Histogram");
    imshow("Histogram", h.GetHistogramImage(src));
    
    waitKey(0);
    return 0;
}

执行结果如下：

image.png

内容来自于书籍。