C++ は K-means アルゴリズムを実装します

原則は省略しますが、コードは次のとおりです。

#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <map>
using namespace std;
#define FLT_MAXX 3.402823466e+38F

class Solution {
    
    
public:
    vector<vector<float>> init_pos(vector<vector<float> >& positions, int k) {
    
    
        vector<vector<float>> ans;
        for(int i=0; i<k; i++) {
    
    
            ans.push_back(positions[i]);
        }
        return ans;
    }
    float cal_len(vector<float>& pos, vector<float>& sta ) {
    
    
        float ret = pow((pos[0] - sta[0])*(pos[0] - sta[0]) + (pos[1] - sta[1])*(pos[1] - sta[1]), 0.5);
        return ret;
    }
    
    vector<int> get_father(vector<vector<float> >& positions, vector<vector<float> >& station, int k) {
    
    
        int n = positions.size();
        vector<int> father(n, 0);
        for(int i=0; i<n; i++) {
    
            // 计算第i个点到第j个站的距离，最小的距离对应的站台标号为fa
            float tmp = FLT_MAXX;
            int fa;
            for(int j=0; j<k; j++) {
    
    
                float len = cal_len(positions[i], station[j]);        // 第i个点和第j个站台的距离
                if(len < tmp) {
    
    
                    tmp = len;
                    fa = j;
                }
            }
            father[i] = fa;
        }
        return father;
        
    }
    void refresh(vector<vector<float> >& positions,vector<vector<float> >& stations, vector<int>& father) {
    
    
        int k = stations.size();
        int n = positions.size();
        map<int, vector<vector<float>>> path;       // 第i个站台所属的所有样本
        for(int i=0; i<n; i++) {
    
    
            float x = positions[i][0];
            float y = positions[i][1];

            int fa = father[i];
            path[fa].push_back(vector<float> {
    
    x, y});
        }

        for(int i=0; i<k; i++) {
    
    
            int len = path[i].size();
            float x = 0;
            float y = 0;
            for(int j=0; j<len; j++) {
    
    
                x += path[i][j][0];
                y += path[i][j][1];
            }
            x /= len;
            y /= len;
            stations[i] = vector<float> {
    
    x, y};
        }

    }
    vector<int> k_means(vector<vector<float> >& positions, int k) {
    
    
        // srand((unsigned int)time(NULL));
        auto station = init_pos(positions, k);
        vector<int> father = get_father(positions, station, k);     // 计算每个点所属的聚类中心
        int epoch = 100;
        for(int i=0; i<epoch; i++) {
    
    
            refresh(positions, station, father);        // 更新station
            father = get_father(positions, station, k);
        }
        
        
        return father;
    }
};

int main()
{
    
    
    vector<vector<float>> positions = {
    
    {
    
    1.5,2.1}, {
    
    0.8,2.1}, {
    
    1.3,2.1}, {
    
    110.5,260.6}, {
    
    21.7, 32.8},{
    
    130.9,150.8},{
    
    32.6,40.7},{
    
    41.5,24.7}};
    Solution base;
    auto father = base.k_means(positions, 3);

    for(int i=0; i<positions.size(); i++)
        cout << father[i] << ' ';


    return 0;
}

// 答案：[1,1,1,0,2,0,2,2]