交换两个不同图像的直方图

Question

我有两个不同的图像（图像A和图像B），其直方图（histImage和histImage1）已经计算出来。 现在我想的是图像A的直方图变成图像B的直方图，使图像B获得相似的图像A.颜色 代码如下：交换直方图

#include "stdafx.h" 
#include "opencv2/highgui/highgui.hpp" 
#include "opencv2/imgproc/imgproc.hpp" 
#include <iostream> 
#include <stdio.h> 

using namespace std; 
using namespace cv; 
int main() 
{ 
    Mat src, dst, src1; 

    /// Load image 
    src = imread("ImageA", 1); // Image A 
    src1 = imread("ImageB", 1); // Image B 

    if(!src.data) 
    { return -1; } 

    /// Separate the image in 3 places (B, G and R) 
    vector<Mat> bgr_planes; 
    vector<Mat> bgr_planes1; 
    split(src, bgr_planes); 
    split(src1, bgr_planes1); 

    /// Establish the number of bins 
    int histSize = 256; 

    /// Set the ranges (for B,G,R)) 
    float range[] = { 0, 256 } ; 
    const float* histRange = { range }; 

    bool uniform = true; bool accumulate = false; 

    Mat b_hist, g_hist, r_hist; //ImageA 
    Mat b_hist1, g_hist1, r_hist1; //ImageB 

    /// Compute the histograms of Image A 
    calcHist(&bgr_planes[0], 1, 0, Mat(), b_hist, 1, &histSize, &histRange, uniform, accumulate); 
    calcHist(&bgr_planes[1], 1, 0, Mat(), g_hist, 1, &histSize, &histRange, uniform, accumulate); 
    calcHist(&bgr_planes[2], 1, 0, Mat(), r_hist, 1, &histSize, &histRange, uniform, accumulate); 
    /// Compute the histograms of Image B 
    calcHist(&bgr_planes1[0], 1, 0, Mat(), b_hist1, 1, &histSize, &histRange, uniform, accumulate); 
    calcHist(&bgr_planes1[1], 1, 0, Mat(), g_hist1, 1, &histSize, &histRange, uniform, accumulate); 
    calcHist(&bgr_planes1[2], 1, 0, Mat(), r_hist1, 1, &histSize, &histRange, uniform, accumulate); 


    // Draw the histograms for B, G and R 
    int hist_w = 512; int hist_h = 400; //Image A 
    int bin_w = cvRound((double) hist_w/histSize); //Image A 
    int hist_w1 = 512; int hist_h1 = 400; //Image B 
    int bin_w1 = cvRound((double) hist_w1/histSize);//Image B 

    Mat histImage(hist_h, hist_w, CV_8UC3, Scalar(0,0,0)); //ImageA 
    Mat histImage1(hist_h1, hist_w1, CV_8UC3, Scalar(0,0,0)); //ImageB 

    /// Normalize the result to [ 0, histImage.rows ] ImageA 
    normalize(b_hist, b_hist, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 
    normalize(g_hist, g_hist, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 
    normalize(r_hist, r_hist, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 
    /// Normalize the result to [ 0, histImage.rows ] ImageB 
    normalize(b_hist1, b_hist1, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 
    normalize(g_hist1, g_hist1, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 
    normalize(r_hist1, r_hist1, 0, histImage.rows, NORM_MINMAX, -1, Mat()); 


    /// Draw for each channel ImageA 
    for(int i = 1; i < histSize; i++) 
    { 
     line(histImage, Point(bin_w*(i-1), hist_h - cvRound(b_hist.at<float>(i-1))) , 
         Point(bin_w*(i), hist_h - cvRound(b_hist.at<float>(i))), 
         Scalar(255, 0, 0), 2, 8, 0 ); 
     line(histImage, Point(bin_w*(i-1), hist_h - cvRound(g_hist.at<float>(i-1))) , 
         Point(bin_w*(i), hist_h - cvRound(g_hist.at<float>(i))), 
         Scalar(0, 255, 0), 2, 8, 0 ); 
     line(histImage, Point(bin_w*(i-1), hist_h - cvRound(r_hist.at<float>(i-1))) , 
         Point(bin_w*(i), hist_h - cvRound(r_hist.at<float>(i))), 
         Scalar(0, 0, 255), 2, 8, 0 ); 
    } 
    //////////////////////////////////////////////////// 
    /// Draw for each channel ImageB 
    for(int i = 1; i < histSize; i++) 
    { 
     line(histImage1, Point(bin_w1*(i-1), hist_h1 - cvRound(b_hist1.at<float>(i-1))) , 
         Point(bin_w1*(i), hist_h1 - cvRound(b_hist1.at<float>(i))), 
         Scalar(255, 0, 0), 2, 8, 0 ); 
     line(histImage1, Point(bin_w1*(i-1), hist_h1 - cvRound(g_hist1.at<float>(i-1))) , 
         Point(bin_w1*(i), hist_h1 - cvRound(g_hist1.at<float>(i))), 
         Scalar(0, 255, 0), 2, 8, 0 ); 
     line(histImage1, Point(bin_w1*(i-1), hist_h1 - cvRound(r_hist1.at<float>(i-1))) , 
         Point(bin_w1*(i), hist_h1 - cvRound(r_hist1.at<float>(i))), 
         Scalar(0, 0, 255), 2, 8, 0 ); 
    } 
    ///////////////////////////////////////////////////// 

    /// Display 
    namedWindow("calcHist", CV_WINDOW_AUTOSIZE); 
    imshow("face ", histImage); //Histogram of Image A 
    /// Display 
    namedWindow("calcHist1", CV_WINDOW_AUTOSIZE); 
    imshow("body ", histImage1); //Histogram of Image B 

    waitKey(0); 

    return 0; 
} 

Answer 1

办法之一是应遵循histogram equalisation中使用的方法。

• 计算直方图（H1和H2）分别为两个图像（I1和I2）和规范他们（在你的代码已经完成）。
• 计算累积直方图 - 也称为累积分布函数 - 对应于H1和H2的C1和C2，如here所述。
• 如here所述，使用累积直方图C2替换I1中每个像素的新值。
• 使用累积直方图C1对I2中的每个像素做同样的处理。