Selecting the pixels with highest intensity in OpenCV - c++

Question

Can anyone help me to find out the top 1% (or say top 100 pixels)brightest pixels with their locations of a gray image in opencv. because cvMinMaxLoc() gives only brightest pixel location.
Any help is greatly appreciated.

Answer 1

this is a simple yet unneficient/stupid way to do it:
for i=1:100
get brightest pixel using cvMinMaxLoc
store location
set it to a value of zero
end
if you don't mind about efficiency this should work.
you should also check cvInRangeS to find other pixels of similar values defining low and high thresholds.

Answer 2

You need to calculate the brightness threshold from the histogram. Then you iterate through the pixels to get those positions that are bright enough to satisfy the threshold. The program below instead applies the threshold to the image and displays the result for demonstration purposes:
#!/usr/bin/env python3
import sys
import cv2
import matplotlib.pyplot as plt
if __name__ == '__main__':
if len(sys.argv) != 2 or any(s in sys.argv for s in ['-h', '--help', '-?']):
print('usage: {} <img>'.format(sys.argv[0]))
exit()
img = cv2.imread(sys.argv[1], cv2.IMREAD_GRAYSCALE)
hi_percentage = 0.01 # we want we the hi_percentage brightest pixels
# * histogram
hist = cv2.calcHist([img], [0], None, [256], [0, 256]).flatten()
# * find brightness threshold
# here: highest thresh for including at least hi_percentage image pixels,
# maybe you want to modify it for lowest threshold with for including
# at most hi_percentage pixels
total_count = img.shape[0] * img.shape[1] # height * width
target_count = hi_percentage * total_count # bright pixels we look for
summed = 0
for i in range(255, 0, -1):
summed += int(hist[i])
if target_count <= summed:
hi_thresh = i
break
else:
hi_thresh = 0
# * apply threshold & display result for demonstration purposes:
filtered_img = cv2.threshold(img, hi_thresh, 0, cv2.THRESH_TOZERO)[1]
plt.subplot(121)
plt.imshow(img, cmap='gray')
plt.subplot(122)
plt.imshow(filtered_img, cmap='gray')
plt.axis('off')
plt.tight_layout()
plt.show()

Answer 3

C++ version based upon some of the other ideas posted:
// filter the brightest n pixels from a grayscale img, return a new mat
cv::Mat filter_brightest( const cv::Mat& src, int n ) {
CV_Assert( src.channels() == 1 );
CV_Assert( src.type() == CV_8UC1 );
cv::Mat result={};
// simple histogram
std::vector<int> histogram(256,0);
for(int i=0; i< int(src.rows*src.cols); ++i)
histogram[src.at<uchar>(i)]++;
// find max threshold value (pixels from [0-max_threshold] will be removed)
int max_threshold = (int)histogram.size() - 1;
for ( ; max_threshold >= 0 && n > 0; --max_threshold ) {
n -= histogram[max_threshold];
}
if ( max_threshold < 0 ) // nothing to do
src.copyTo(result);
else
cv::threshold(src, result, max_threshold, 0., cv::THRESH_TOZERO);
return result;
}
Usage example: get top 1%
auto top1 = filter_brightest( img, int((img.rows*img.cols) * .01) );

Answer 4

Try using cvThreshold instead.

Answer 5

Well the most logical way is to iterate the whole picture, then get the max and min value of the pixels.
Then chose a threshold which will give you the desired percent(1% in your case).
After that iterate again and save the i and j coordinates of each pixel above the given threshold.
This way you'll iterate the matrix only two times instead of 100(or 1% of the pixels times) and choosing the brightest and deleting it.
OpenCV mats are multidimensional arrays. Gray image is two dimensional array with values from 0 to 255.
You can iterate trough the matrix like this.
for(int i=0;i < mat.height();i++)
for(int j=0;j < mat.width();j++)
mat[i][j];