I tried to get horizontal projection using countNonZero() function as below.
Mat src = imread(INPUT_FILE, CV_LOAD_IMAGE_COLOR);
Mat binaryImage = src.clone();
cvtColor(src, src, CV_BGR2GRAY);
Mat horizontal = Mat::zeros(1,binaryImage.cols, CV_8UC1);
for (int i = 0; i<binaryImage.cols; i++)
{
Mat roi = binaryImage(Rect(0, 0, 1, binaryImage.rows));
horizontal.at<int>(0,i) = countNonZero(roi);
cout << "Col no:" << i << " >>" << horizontal.at<int>(0, i);
}
But an error is occured in the line of calling countonZero() function. Error is as follows.
OpenCV Error: Assertion failed (src.channels() == 1 && func != 0) in cv::countNo
nZero, file C:\builds\2_4_PackSlave-win32-vc12-shared\opencv\modules\core\src\st
at.cpp, line 549
Can somebody please point out the mistake?
Assertion src.channels() == 1
means that image should have 1 channel, i.e. it has to be gray, not colored. You are calling countNonZero
on roi
, which is a subimage of binaryImage
, which is a clone of src
, which is originally colored.
I suppose you wanted to write cvtColor(binaryImage, binaryImage, CV_BGR2GRAY);
. In this case it makes sense. However, I do not see you using src
anywhere again, so perhaps you do not need this intermediate image. In case you do, do not call "binary", since "binary" in computer vision usually stands for black-or-white image, only two colors. Your image is "gray", since it has all shades of black and white.
Concerning your original task, Miki is right, you should use cv::reduce
for it. He already gave you an example on how to use it.
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