根據(jù)《統(tǒng)計(jì)學(xué)習(xí)方法》以例4.1的數(shù)據(jù)為例實(shí)現(xiàn)的樸素貝葉斯。

感覺最后計(jì)算比較時(shí)候可以避免使用double，但是為了思路清晰就這樣把。
#include<iostream>
#include<map>
using namespace std;

int X1[]{1,1,1,1,1,2,2,2,2,2,3,3,3,3,3};
char X2[]{'S','M','M','S','S','S','M','M','L','L','L','M','M','L','L'};
int Y[]{-1,-1,1,1,-1,-1,-1,1,1,1,1,1,1,1,-1};

int main() {
    int x1;
    char x2{};
    cin >> x1;
    cin >> x2;
    map<string, int> ycount;//記錄y =1 -1的次數(shù)
    map<string, int> x1count1;//記錄y=1,-1 ，x1匹配的次數(shù)
    map<string, int> x2count1;//記錄y=1,-1 ，x2匹配的次數(shù)
    for (int i = 0; i < sizeof(Y) / sizeof(Y[0]); i++) {
        if (Y[i] == 1) {
            ycount["y=1"] ++;
            if (X1[i] == x1) {
                x1count1["y=1"] ++;
            }
            if (X2[i] == x2) {
                x2count1["y=1"] ++;
            }
        }
        else {
            ycount["y=-1"] ++;
            if (X1[i] == x1) {
                x1count1["y=-1"] ++;
            }
            if (X2[i] == x2) {
                x2count1["y=-1"] ++;
            }
        }
    }
    //計(jì)算概率
    double rs1 = (double)ycount["y=1"]/15 * (double)x1count1["y=1"]/ (double)ycount["y=1"] * (double)x2count1["y=1"]/ (double)ycount["y=1"];
    double rs2 = (double)ycount["y=-1"] /15 * (double)x1count1["y=-1"] / (double)ycount["y=-1"] * (double)x2count1["y=-1"] / (double)ycount["y=-1"];
            cout << rs1<<endl;
    cout << rs2<<endl;
           (rs1 > rs2) ? cout << "y=1" : cout << "y=-1";
    system("pause");
}