下面是mkh答案的一个变体，比它运行得快得多，还有scipy.stats。皮尔逊，使用numba。

import numba

@numba.jit
def corr(data1, data2):
    M = data1.size

    sum1 = 0.
    sum2 = 0.
    for i in range(M):
        sum1 += data1[i]
        sum2 += data2[i]
    mean1 = sum1 / M
    mean2 = sum2 / M

    var_sum1 = 0.
    var_sum2 = 0.
    cross_sum = 0.
    for i in range(M):
        var_sum1 += (data1[i] - mean1) ** 2
        var_sum2 += (data2[i] - mean2) ** 2
        cross_sum += (data1[i] * data2[i])

    std1 = (var_sum1 / M) ** .5
    std2 = (var_sum2 / M) ** .5
    cross_mean = cross_sum / M

    return (cross_mean - mean1 * mean2) / (std1 * std2)

Pearson coefficient calculation using pandas in python: I would suggest trying this approach since your data contains lists. It will be easy to interact with your data and manipulate it from the console since you can visualise your data structure and update it as you wish. You can also export the data set and save it and add new data out of the python console for later analysis. This code is simpler and contains less lines of code. I am assuming you need a few quick lines of code to screen your data for further analysis

例子:

data = {'list 1':[2,4,6,8],'list 2':[4,16,36,64]}

import pandas as pd #To Convert your lists to pandas data frames convert your lists into pandas dataframes

df = pd.DataFrame(data, columns = ['list 1','list 2'])

from scipy import stats # For in-built method to get PCC

pearson_coef, p_value = stats.pearsonr(df["list 1"], df["list 2"]) #define the columns to perform calculations on
print("Pearson Correlation Coefficient: ", pearson_coef, "and a P-value of:", p_value) # Results 

但是，在分析之前，你没有发布你的数据给我看数据集的大小或可能需要的转换。

计算相关:

相关性-衡量两个不同变量的相似性

使用皮尔逊相关

from scipy.stats import pearsonr
# final_data is the dataframe with n set of columns
pearson_correlation = final_data.corr(method='pearson')
pearson_correlation
# print correlation of n*n column

使用斯皮尔曼相关

from scipy.stats import spearmanr
# final_data is the dataframe with n set of columns
spearman_correlation = final_data.corr(method='spearman')
spearman_correlation
# print correlation of n*n column

使用Kendall相关

kendall_correlation=final_data.corr(method='kendall')
kendall_correlation

一个替代方法可以是一个来自linreturn的本地scipy函数，它计算:

斜率:回归线的斜率 截距:回归线的截距 R-value:相关系数 p值:零假设为斜率为零的假设检验的双面p值 stderr:估计的标准错误

这里有一个例子:

a = [15, 12, 8, 8, 7, 7, 7, 6, 5, 3]
b = [10, 25, 17, 11, 13, 17, 20, 13, 9, 15]
from scipy.stats import linregress
linregress(a, b)

会回复你:

LinregressResult(slope=0.20833333333333337, intercept=13.375, rvalue=0.14499815458068521, pvalue=0.68940144811669501, stderr=0.50261704627083648)

如果你不喜欢安装scipy，我使用了这个快速的hack，稍微修改了Programming Collective Intelligence:

def pearsonr(x, y):
  # Assume len(x) == len(y)
  n = len(x)
  sum_x = float(sum(x))
  sum_y = float(sum(y))
  sum_x_sq = sum(xi*xi for xi in x)
  sum_y_sq = sum(yi*yi for yi in y)
  psum = sum(xi*yi for xi, yi in zip(x, y))
  num = psum - (sum_x * sum_y/n)
  den = pow((sum_x_sq - pow(sum_x, 2) / n) * (sum_y_sq - pow(sum_y, 2) / n), 0.5)
  if den == 0: return 0
  return num / den

Pearson相关性可以用numpy的corrcoef来计算。

import numpy
numpy.corrcoef(list1, list2)[0, 1]