皮尔逊相关系数R的代码实现

皮尔逊相关系数代码实现

①直接用numpy的corrcoef方法

from math import sqrt
import numpy as np

x = [2,7,18,88,157, 90,177,570]
y = [3,5,15,90,180, 88,160,580]

print (np.corrcoef(x,y))  # ✔

输出结果—— （右上左下为R）

[[1. 0.99834875]

[0.99834875 1. ]]

②调用scipy库函数

from scipy.stats import pearsonr

v1 = [2,7,18,88,157, 90,177,570]
v2 = [3,5,15,90,180, 88,160,580]

pccs = pearsonr(v1, v2)
print(pccs)

输出结果—— （第一位为R）

(0.99834874864405, 1.1241947891650549e-08)

②自己手动实现

import math
import numpy as np

def pearson(v1, v2):
    n = len(v1)
    #simple sums
    sum1 = sum(float(v1[i]) for i in range(n))
    sum2 = sum(float(v2[i]) for i in range(n))
    #sum up the squares
    sum1_pow = sum([pow(v, 2.0) for v in v1])
    sum2_pow = sum([pow(v, 2.0) for v in v2])
    #sum up the products
    p_sum = sum([v1[i] * v2[i] for i in range(n)])
    #分子num，分母denominator
    num = p_sum - (sum1*sum2/n)
    den = math.sqrt((sum1_pow-pow(sum1, 2)/n)*(sum2_pow-pow(sum2, 2)/n))
    if den == 0:
        return 0.0
    return num/den

vector1 = [2,7,18,88,157, 90,177,570]
vector2 = [3,5,15,90,180, 88,160,580]

print(pearson(vector2,vector1))

输出结果——

0.9983487486440501