The following piece of code does not seem to work. There is a similar question here How to fill Matrix with zeros in OpenCV? but, I want a 1-D mat, not a 2-D mat as mentioned in this link.
int histSize = 8;
Mat colorHist;
for (int i = 0; i < (histSize * histSize * histSize); i++)
colorHist.at<float>(i) = 0.0;
You may try something like that:
1 - First you create a float array which is your 1-D data structure:
float arr[10] = {0}; // initialize it to all 0`s
2 - Now create your opencv matrix as follows and fill the array into it:
cv::Mat colorHist = cv::Mat(2, 4, CV_32F, arr);
3 - if you want to access individual entries use something like:
for(int i=0; i<10; i++) {
colorHist.at<float>(0,i);
}
where i is the index of the entry you want from 0 to 9.
or just:
colorHist.at<float>(0,2);
if you need individually. here we get the entry with index 2 which will return 0 of course since the matrix is all zeros at that point.
EDIT: As Nicholas suggested:
cv::Mat colorHist = cv::Mat::zeros(1, 10, CV_32F); // size=10
is a shorter way of creating a 1-D row matrix with all zeros if you do not want to deal with a float array (credits go to Nicholas). and access is of course:
for(int i=0; i<10; i++) {
colorHist.at<float>(i);
}
Hope that helps!
In the code you have posted, you have not initialized the Mat to any size or type. However, there is a very easy way to initialize a matrix with zeros.
cv::Mat colorHist = cv::Mat::zeros(1, histSize, CV_32F);
This will produce a row vector (1-dimensional) with histSize columns, CV_32F simply refers to the data type the matrix is dealing with (float in this case).
If you wish to work with column vectors instead, simply swap the dimensions:
cv::Mat colorHist = cv::Mat::zeros(histSize, 1, CV_32F);
for(int i = 0; i < histSize; i++)
float c = colorHist.at<float>(i);
