Mahout-Pearson correlation的实现

大家好，又见面了，我是全栈君。

计算公式：

并通过以下代码对Mahout in Action的结果进行了验证：

代码例如以下： ` package com.example.mahout;

public class TestColl {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    //int a[]={68,71,62,75,58,60,67,68,71,69,68,67,63,62,60,63,65,67,63,61};
    //double b[] ={4.1,4.6,3.8,4.4,3.2,3.1,3.8,4.1,4.3,3.7,3.5,3.2,3.7,3.3,3.4,4.0,4.1,3.8,3.4,3.6};
    double a[]={5,3.0,2.5};
    double b[]={4,3,2.0};

    int i,j;
    double sum_a=0,sum_b=0,sum_XY=0,sum_X=0,sum_Y=0,sum_X2=0,sum_Y2=0;
    double  mean_a,var_a,mean_b,var_b;
    for(i=0;i<a.length;i++){
        sum_a+=a[i];
        sum_b+=b[i];
        sum_XY+=a[i]*b[i];
        sum_X2+=a[i]*a[i];
        sum_Y2+=b[i]*b[i];
    }

    mean_a = sum_a/a.length;
    mean_b = sum_b/b.length;
    System.out.println("sum_a:"+sum_a);
    System.out.println("sum_b:"+sum_b);
    System.out.println("mean_a:"+mean_a);
    System.out.println("mean_b:"+mean_b);
    sum_X=sum_a;
    sum_Y = sum_b;

    sum_a=sum_b=0;
    for(i=0;i<a.length;i++){
        sum_a+=(a[i]-mean_a)*(a[i]-mean_a);
        sum_b+=(b[i]-mean_b)*(b[i]-mean_b);
    }
    var_a=sum_a/(a.length-1);
    var_b=sum_b/(a.length-1);
    System.out.println("var_a:"+var_a);
    System.out.println("var_b:"+var_b);
    System.out.println("sum_XY:"+sum_XY);
    System.out.println("sum_X:"+sum_X);
    System.out.println("sum_X2:"+sum_X2);
    System.out.println("sum_Y2:"+sum_Y2);
    double r_up = a.length*sum_XY-sum_X*sum_Y;
    double r_down = Math.sqrt((a.length*sum_X2-sum_X*sum_X)*(a.length*sum_Y2-sum_Y*sum_Y));
    double r=r_up/r_down;

    System.out.println("r_up:"+r_up);
    System.out.println("r_down:"+r_down);
    System.out.println("r:"+r);

}

复制

} `

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