OpenCV3.2图像分割 实例2：KMeans对随机生成数据进行分类

 1 #include <opencv2/opencv.hpp>
 2 #include <iostream>
 3  
 4 using namespace cv;
 5 using namespace std;
 6  
 7 int main(int argc, char** argv) {
 8     Mat img(500, 600, CV_8UC3);//定义一张图
 9     RNG rng(12345);//定义随机数
10     //不同类定义为不同颜色
11     Scalar colorTab[] = {
12         Scalar(0, 0, 255),
13         Scalar(0, 255, 0),
14         Scalar(255, 0, 0),
15         Scalar(0, 255, 255),
16         Scalar(255, 0, 255)
17     };
18  
19     int numCluster = rng.uniform(2, 5);//定义分类种类数量块
20     printf("number of clusters : %d\n", numCluster);
21     //设置从原图像中抽取多少个数据点
22     int sampleCount = rng.uniform(5, 1000);
23     Mat points(sampleCount, 1, CV_32FC2);
24     Mat labels;
25     Mat centers;
26  
27     // 生成随机数
28     for (int k = 0; k < numCluster; k++) {
29         Point center;
30         center.x = rng.uniform(0, img.cols);
31         center.y = rng.uniform(0, img.rows);
32         //得到不同小块
33         Mat pointChunk = points.rowRange(k*sampleCount / numCluster, 
34             k == numCluster - 1 ? sampleCount : (k + 1)*sampleCount / numCluster);
35         //用随机数对小块点进行填充
36         rng.fill(pointChunk, RNG::NORMAL, Scalar(center.x, center.y), Scalar(img.cols*0.05, img.rows*0.05));
37     }
38     randShuffle(points, 1, &rng);
39  
40     // 使用KMeans
41     kmeans(points, numCluster, labels, TermCriteria(TermCriteria::EPS + TermCriteria::COUNT, 10, 0.1), 3, KMEANS_PP_CENTERS, centers);
42  
43     // 用不同颜色显示分类
44     img = Scalar::all(255);
45     for (int i = 0; i < sampleCount; i++) {
46         int index = labels.at<int>(i);
47         Point p = points.at<Point2f>(i);
48         circle(img, p, 2, colorTab[index], -1, 8);
49     }
50  
51     // 每个聚类的中心来绘制圆
52     for (int i = 0; i < centers.rows;  i++) {
53         int x = centers.at<float>(i, 0);
54         int y = centers.at<float>(i, 1);
55         printf("c.x= %d, c.y=%d", x, y);
56         circle(img, Point(x, y), 40, colorTab[i], 1, LINE_AA);
57     }
58  
59     imshow("KMeans-Data-Demo", img);
60     waitKey(0);
61     return 0;
62 }