皮尔逊相关系数的计算(python代码版)

from math import sqrt

def multipl(a,b):
    sumofab=0.0
    for i in range(len(a)):
        temp=a[i]*b[i]
        sumofab+=temp
    return sumofab

def corrcoef(x,y):
    n=len(x)
    #求和
    sum1=sum(x)
    sum2=sum(y)
    #求乘积之和
    sumofxy=multipl(x,y)
    #求平方和
    sumofx2 = sum([pow(i,2) for i in x])
    sumofy2 = sum([pow(j,2) for j in y])
    num=sumofxy-(float(sum1)*float(sum2)/n)
    #计算皮尔逊相关系数
    den=sqrt((sumofx2-float(sum1**2)/n)*(sumofy2-float(sum2**2)/n))
    return num/den

x = [0,1,0,3]
y = [0,1,1,1]

print corrcoef(x,y) #0.471404520791

  

posted @ 2015-08-27 17:40  dayday+up  阅读(25566)  评论(1编辑  收藏  举报