I have B,G,R histograms that look like the following: Image Histogram for B channel of an image
Description: On the X axis, I have the values from 0-255, that each pixel ranges from, and on Y axis, I have the number of pixels that have that particular X value.
My code for the same is:
hist1 = cv2.calcHist([image],[0],None,[256],[0,256])
plt.plot(hist1, color='b')
plt.xlabel("Value (blue)")
plt.ylabel("Number of pixels")
plt.title('Image Histogram For Blue Channel')
plt.show()
My question is, that I need to get the same plot - X axis with values, and Y axis with number of pixels, for HSV channels. Basically, instead of B, G, and R plots, I need the same histogram, but one that gets H, S, I.
I got the following code:
img2 = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)
h, s, v = img2[:,:,0], img2[:,:,1], img2[:,:,2]
hist_h = cv2.calcHist([h],[0],None,[256],[0,256])
#hist_s = cv2.calcHist([s],[0],None,[256],[0,256])
#hist_v = cv2.calcHist([v],[0],None,[256],[0,256])
plt.plot(hist_h, color='r', label="hue")
Which gives me the following plot: Hue plot for an image
But from what I've read so far, BGR and HSV are different color spaces. So, I want to know, that when I'm using the calcHist function after converting to HSV, and then splitting into three channels, those channels by default are H,S and V? It's not that they're actually only BGR, but just simply mislabelled H, S and V? I just want to verify how both the methods are practically the same, but BGR and HSV are different color spaces.
Edit: Here's the source image Image
CodePudding user response:
Most likely you have a synthetic image with nothing in the red and green channels and some random data centred on 128 in the blue channel.
When you go to HSV colourspace, all the hues are centred on 110 which corresponds to 220 degrees which is blue in the regular 0..360 HSV range. Remember OpenCV uses a range of 0..180 for Hue when using uint8 so that it fits in uint8's range of 0..255. So you effectively need to multiply the 110 you see in your Hue histogram by 2... making 220 which is blue.
See bottom part of this figure.
CodePudding user response:
The OpenCV documentation shows you how to do this for HSV. The one thing to take into account is that while in RGB space each channel has a range of 0 to 255. In HSV space each channel is not te same.
The code below is a copy paste from the documentation and reference link below.
#include <opencv2/imgproc.hpp>
#include <opencv2/highgui.hpp>
using namespace cv;
int main( int argc, char** argv )
{
Mat src, hsv;
if( argc != 2 || !(src=imread(argv[1], 1)).data )
return -1;
cvtColor(src, hsv, COLOR_BGR2HSV);
// Quantize the hue to 30 levels
// and the saturation to 32 levels
int hbins = 30, sbins = 32;
int histSize[] = {hbins, sbins};
// hue varies from 0 to 179, see cvtColor
float hranges[] = { 0, 180 };
// saturation varies from 0 (black-gray-white) to
// 255 (pure spectrum color)
float sranges[] = { 0, 256 };
const float* ranges[] = { hranges, sranges };
MatND hist;
// we compute the histogram from the 0-th and 1-st channels
int channels[] = {0, 1};
calcHist( &hsv, 1, channels, Mat(), // do not use mask
hist, 2, histSize, ranges,
true, // the histogram is uniform
false );
double maxVal=0;
minMaxLoc(hist, 0, &maxVal, 0, 0);
int scale = 10;
Mat histImg = Mat::zeros(sbins*scale, hbins*10, CV_8UC3);
for( int h = 0; h < hbins; h )
for( int s = 0; s < sbins; s )
{
float binVal = hist.at<float>(h, s);
int intensity = cvRound(binVal*255/maxVal);
rectangle( histImg, Point(h*scale, s*scale),
Point( (h 1)*scale - 1, (s 1)*scale - 1),
Scalar::all(intensity),
-1 );
}
namedWindow( "Source", 1 );
imshow( "Source", src );
namedWindow( "H-S Histogram", 1 );
imshow( "H-S Histogram", histImg );
waitKey();
}
Reference Documentation Link: https://docs.opencv.org/3.4/d6/dc7/group__imgproc__hist.html#ga4b2b5fd75503ff9e6844cc4dcdaed35d