OpenCV: Kmeans的使用一维和二维点集

OpenCVKmeans算法默认使用了Kmeans++选取种子点

参考:OpenCv中Kmeans算法实现和使用

//效果:根据半径聚类,并不一定能得到好的结果。
float CBlotGlint::ClusterByR(  )
{
	//根据半径大小聚类,找出合适的类别个数和每一类的个数
	std::vector<float>   radiuses(this->blobs.size() );
	std::vector<std::pair<float,int> >   radiusesIdx(this->blobs.size() );
	for ( int i=0; i< this->blobs.size(); ++i ){
		radiuses[i] =this->blobs[i].diaGlint;
		radiusesIdx[i].first  = radiuses[i];
		radiusesIdx[i].second = i;
	}

	{
		using namespace cv;

		this->blobs[0].diaGlint;

		const int MAX_CLUSTERS = 5;
		Scalar colorTab[] =
		{
			Scalar(0, 0, 255),
			Scalar(0,255,0),
			Scalar(255,100,100),
			Scalar(255,0,255),
			Scalar(0,255,255)
		};

		Mat img( 500, 500, CV_8UC3 );
		RNG rng( 12345 );
		std::vector<std::vector<float> >  outC;
		std::vector<std::vector<std::pair<float,int> > >  outCidx;

		{
			outC.clear();
			int k;
			int clusterCount = rng.uniform(2, MAX_CLUSTERS+1);
			int i;
			int sampleCount =radiuses.size();// rng.uniform(1, 1001);
			Mat points(sampleCount, 1, CV_32FC1), labels;
			for ( int i=0; i< sampleCount; ++i ){
				points.at<float>(i) = radiuses[i];
			}

			clusterCount = MIN(clusterCount, sampleCount);
			clusterCount = std::max(clusterCount,3);
			clusterCount = 3;

			Mat centers;
			kmeans(points, clusterCount, labels,
				TermCriteria( CV_TERMCRIT_EPS+CV_TERMCRIT_ITER, 10, 1.0),
				3, KMEANS_PP_CENTERS, centers);

			outC.resize( clusterCount );
			outCidx.resize( clusterCount );

			img = Scalar::all(0);
			for( i = 0; i < sampleCount; i++ )
			{
				int clusterIdx = labels.at<int>( i );
				outC[clusterIdx].push_back( points.at<float>( i  ) );
				outCidx[clusterIdx].push_back( 
					std::make_pair ( points.at<float>( i  ) ,radiusesIdx[i].second ) );

				Point ipt = this->blobs[i].centerOfGlint;
				circle( img, ipt, 2, colorTab[clusterIdx], CV_FILLED, CV_AA );
			}

			cv::imshow("clusters", img);

			char key = (char)cv::waitKey(1);
		}

		return 0;

	}

	return 1.0;
}

即使如此,每次聚类的效果仍然不一定相同,显示一定的随机性。



posted @ 2017-08-19 23:28  wishchin  阅读(510)  评论(0编辑  收藏  举报