【皮尔逊相关系数】类似于余弦定理的使用(推荐),直接上代码

# 也是用于内容推荐


def pearson(p,q):
     只计算两者共同有的
    same = 0
    for i in p:
        if i  q:
            same += 1

    n = same
     分别求p，q的和
    sumx = sum([p[i]  range(n)])
    sumy = sum([q[i]  range(n)])
     分别求出p，q的平方和
    sumxsq = sum([p[i] ** 2  range(n)])
    sumysq = sum([q[i] ** 2  求出p，q的乘积和
    sumxy = sum([p[i] * q[i]  print sumxy
     求出pearson相关系数
    up = sumxy - sumx * sumy / n
    down = ((sumxsq - pow(sumxsq,2) / n) * (sumysq - pow(sumysq,2) / n)) ** .5
     若down为零则不能计算，return 0
    if down == 0: return 0
    r = up / down
     r



p = [0,1,1]
q = [0,1)">]
print (pearson(p,q))

# 得出的结果是1.0