计算opencv二进制图像中的“白色”像素（有效）

Question

I am trying to count all the white pixels in an OpenCV binary image. My current code is as follows:

  whitePixels = 0;
  for (int i = 0; i < height; ++i)
    for (int j = 0; j < width; ++j)
      if (binary.at<int>(i, j) != 0)
        ++whitePixels;

However, after profiling with gprof I've found that this is a very slow piece of code, and a large bottleneck in the program.

Is there a method which can compute the same value faster?

Answer 1

cv::CountNonZero . Usually the OpenCV implementation of a task is heavily optimized.

Answer 2

You can use paralell computing. You divide the image in N parts and run your code in differents threads then you get the result of each threads and after this you can add this results for obtain the finally amount.

Answer 3

Actually binary.at<int>(i, j) is slow access!

Here is simple code that access faster than yours.

for (int i = 0; i < height; ++i)
{
uchar * pixel = image.ptr<uchar>(i);
    for (int j = 0; j < width; ++j)
{
  if(pixel[j]!=0)
   {
      //do your job
   }
}
}

Answer 4

The last pixel in a row is usually followed by the first pixel in the next row (C code):

limit=width*height;
i=0;
while (i<limit)
{
  if (binary.at<int>(0,i) != 0) ++whitePixels;
  ++i;
}