我正试图从图像中计算对象.我使用日志照片,并使用一些步骤来获取二进制图像.
这是我的代码:
#include <opencv2/core/core.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>
#include <features2d.hpp>
using namespace cv;
using namespace std;
int main(int argc, char *argv[])
{
//load image
Mat img = imread("kayu.jpg", CV_LOAD_IMAGE_COLOR);
if(img.empty())
return -1;
//namedWindow( "kayu", CV_WINDOW_AUTOSIZE );
imshow("kayu", img);
//convert to b/w
Mat bw;
cvtColor(img, bw, CV_BGR2GRAY);
imshow("bw1", bw);
threshold(bw, bw, 40, 255, CV_THRESH_BINARY);
imshow("bw", bw);
//distance transform & normalisasi
Mat dist;
distanceTransform(bw, dist, CV_DIST_L2, 3);
normalize(dist, dist, 0, 2., NORM_MINMAX);
imshow("dist", dist);
//threshold to draw line
threshold(dist, dist, .5, 1., CV_THRESH_BINARY);
imshow("dist2", dist);
//dist = bw;
//dilasi
Mat dilation, erotion, element;
int dilation_type = MORPH_ELLIPSE;
int dilation_size = 17;
element = getStructuringElement(dilation_type, Size(2*dilation_size + 1, 2*dilation_size+1), Point(dilation_size, dilation_size ));
erode(dist, erotion, element);
int erotionCount = 0;
for(int i=0; i<erotionCount; i++){
erode(erotion, erotion, element);
}
imshow("erotion", erotion);
dilate(erotion, dilation, element);
imshow("dilation", dilation);
waitKey(0);
return 0;
}
如您所见,我使用侵蚀和膨胀来获得更好的圆形对象.我的问题是,我坚持计算对象.我尝试了SimpleBlobDetector,但我没有得到任何东西,因为当我尝试将“dilation”步骤的结果转换为CV_8U时,白色对象消失了.我使用findContours()时也遇到了错误.它讲述了关于图像通道的一些信息.我无法在这里显示错误,因为这个步骤太多了,我已经从代码中删除了它.
顺便说一句,最后,我得到了一个图像通道.
我可以用它来计算,或者我必须转换它,最好的方法是什么?
最佳答案 两个简单的步骤:
>查找二值化图像的轮廓.
>获取轮廓的数量.
码:
int count_trees(const cv::Mat& bin_image){
cv::Mat img;
if(bin_image.channels()>1){
cv::cvtColor(bin_image,img,cv::COLOR_BGR2GRAY);
}
else{
img=bin_image.clone();;
}
if(img.type()!=CV_8UC1){
img*=255.f; //This could be stupid, but I do not have an environment to try it
img.convertTo(img,CV_8UC1);
}
std::vector<std::vector<cv::Point>> contours
std::vector<Vec4i> hierarchy;
cv::findContours( img, contours, hierarchy, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);
return contours.size();
}