计数黑白图像的像素（OpenCV / C ++）

Question

I am a beginner to opencv / C ++ and would like your help with a problem that seems simple. As an example, I have this image:

... And would disregard the background, which will always be white, and the image is always black and white, leaving only the cloud to be able to count three things:

1. The number of pixels of the figure (cloud only, disregarding the background).
2. The number of white pixels. (cloud only)
3. The number of black pixels. (cloud only)

I know that to achieve 2, with a subtraction with it the 3rd.

Thanks!

Answer 1

Here's a very simple method

• Count the black pixels (3)
• Flood-fill with black starting at any corner
• Count the number of remaining white pixels (2)
• Add results (2) and (3) to get (1)

Answer 2

You can use this code to count to find black pixel in an image.Apply same logic for counting white pixels.

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

using namespace std;
using namespace cv;

int main()
{

Mat img = imread("/home/jordan/opencv/Test/edge_detection/img2.bmp",0);
cout<<"cols= "<<img.cols<<"row= "<<img.rows<<"\n";
int i,j,count=0;
for(i=0;i<img.cols;i++)
{
for(j=0;j<img.rows;j++)
{
int k=img.at<uchar>(j,i);
if(k==0)
{  
count++;
cout<<"col="<<i<<"row="<<j<<"k= "<< k<<"\t\n";
}
}
} 
cout<<"count="<<count<<"\n";
}

Answer 3

First of all, you can use cv::countNonZero(img) to count non-black pixels. Thus you can use (img.rows * img.cols) - cv::countNonZero(img) to count black pixels.

Now, try the following: First, count the black pixels. Then try to use morphological opening to remove those white blobs inside the clould. Afterwards, count the black pixels again. Now you can substract the first count from the second one and you'll get the amount of white pixels inside those blobs.

However, this method will not be accurate, because the morphological opening will also alter the cloud a bit on the outside. An alternative would be the method suggested in an other answer, to use flood fill to fill the surrounding white pixels with black.

Morphological opening works like this by the way:

cv::Mat element = cv::getStructuringElement(cv::MORPH_ELLIPSE, cv::Size(2 * openingRadius + 1, 2 * openingRadius + 1));
cv::morphologyEx(mat, mat, cv::MORPH_OPEN, element);

Or, with cuda:

cv::Mat element = cv::getStructuringElement(cv::MORPH_ELLIPSE, cv::Size(2 * openingRadius + 1, 2 * openingRadius + 1));
cv::cuda::createMorphologyFilter(cv::MORPH_OPEN, CV_8UC1, element)->apply(gpuMat, gpuMat);

Answer 4

All the tools you need are cv::findContour and cv::drawContour . First of all you need to get rid of the white background, then do your jobs with the cloud:

findContours(the_image, contours, hierarchy, CV_RETR_LIST , CV_CHAIN_APPROX_NONE );

from contours, find the second big vector(contour) and that is the cloud's boundry. The biggest one by the way, would be the outer contour of the white background. Then...

cv::Mat test_img = cv::Mat::zeros(the_img.size(),CV_8U);
drawContours(test_img , contours, contourIdx, Scalar(255,255,255), CV_FILLED ,8);

contourIdx is the index of that second big contour. Now in test_img you have a cloud that is white completely. Now use countNonZero to get the number of cloud's pixels(num1). In further you have to multiply this image (as a mask) to the original image and then you have the original image but without the white background. Use countNonZero again to have the number of the white pixels in the cloud(num2). And finally num1-num2 gives you the number of black pixels in the cloud. Comment for me if you had trouble with findContur .