如何在c++中平平颜色直方图？

Question

嗨，我用python编写了以下代码行：

# convert the image to HSV color-space
image = cv2.cvtColor(image, cv2.COLOR_BGR2HSV)

# compute the color histogram
hist  = cv2.calcHist([image], [0, 1, 2], None, [bins, bins, bins], [5, 240, 5, 240, 5, 240])

# normalize the histogram
cv2.normalize(hist, hist)

# return the histogram
return hist.flatten()

我现在正试图用c++重写它。我在calculation.html找到了一个很好的例子

我现在面临的问题是在c++中扁平hist，例如在python中，code.This是python (512，)中扁平hist输出的形状。对于如何在c++中获得相同的结果，有什么想法吗？

(编辑)到目前为止的c++代码。

大小(500,500)；图像= imread("C:\johan.jpg"，IMREAD_COLOR)；

 resize(image,image,size);//resize image

 cvtColor(image, image, CV_BGR2HSV);
// Separate the image in 3 places ( H, S and V )
 vector<Mat> bgr_planes;
 split(image, bgr_planes );

 vector<Mat> hist_flat;

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

 // Set the ranges ( for H,S,V) )
 float range[] = {5, 240} ;
 const float* histRange = { range };

 bool uniform = true; bool accumulate = false;

 Mat b_hist, g_hist, r_hist;

 cout << " Working fine Johan...";

 // Compute the histograms:
 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 );
 //calcHist( &image,3, 0, Mat(), hist_flat, 1, &histSize, &histRange, uniform, accumulate );



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



 Mat histImage(hist_h,hist_w, CV_8UC3, Scalar(0,0,0));

 // Normalize the result to [ 0, histImage.rows ]
 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() );



 // Draw for each channel
 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  );
 }

 // Display

 imshow("calcHist Demo", histImage );
 imshow("The image resized",image);

Answer 1

再给这个问题加上一个答案。由于您使用的是OpenCV cv::Mat作为直方图持有者，因此将其扁平化的一种方法是使用reshape (例如：

// create mat a with 512x512 size and float type
cv::Mat a(512,512,CV_32F);
// resize it to have only 1 row
a = a.reshape(0,1);

这个O(1)函数不复制元素，只需将cv::Mat标头更改为具有正确的大小。

之后，您将有一个包含262144列的1行cv::mat。

Answer 2

基本上，您想要扁平一个2D数组( hist = cv2.calcHist([image], [0, 1, 2], None, [bins, bins, bins], [5, 240, 5, 240, 5, 240])是2D数组235x3 )

这方面最简单的代码在function in C++ similar to numpy flatten中

基本算法是( cf http://www.ce.jhu.edu/dalrymple/classes/602/Class12.pdf )。

for (q = 0; q < n; q++)
{
    for (t = 0; t < m; t++)
    {
        b[q * n + t] = a[q][t];  <-------
    }
}

来源：C++ 2D array to 1D array

(用于3D数组cf How to "flatten" or "index" 3D-array in 1D array? )