这是使用numpy的Pearson Correlation函数的实现:

def corr(data1, data2):
    "data1 & data2 should be numpy arrays."
    mean1 = data1.mean() 
    mean2 = data2.mean()
    std1 = data1.std()
    std2 = data2.std()

#     corr = ((data1-mean1)*(data2-mean2)).mean()/(std1*std2)
    corr = ((data1*data2).mean()-mean1*mean2)/(std1*std2)
    return corr

你可以看看scipy.stats:

from pydoc import help
from scipy.stats.stats import pearsonr
help(pearsonr)

>>>
Help on function pearsonr in module scipy.stats.stats:

pearsonr(x, y)
 Calculates a Pearson correlation coefficient and the p-value for testing
 non-correlation.

 The Pearson correlation coefficient measures the linear relationship
 between two datasets. Strictly speaking, Pearson's correlation requires
 that each dataset be normally distributed. Like other correlation
 coefficients, this one varies between -1 and +1 with 0 implying no
 correlation. Correlations of -1 or +1 imply an exact linear
 relationship. Positive correlations imply that as x increases, so does
 y. Negative correlations imply that as x increases, y decreases.

 The p-value roughly indicates the probability of an uncorrelated system
 producing datasets that have a Pearson correlation at least as extreme
 as the one computed from these datasets. The p-values are not entirely
 reliable but are probably reasonable for datasets larger than 500 or so.

 Parameters
 ----------
 x : 1D array
 y : 1D array the same length as x

 Returns
 -------
 (Pearson's correlation coefficient,
  2-tailed p-value)

 References
 ----------
 http://www.statsoft.com/textbook/glosp.html#Pearson%20Correlation

你可以用pandas.DataFrame这样做。相关系数:

import pandas as pd
a = [[1, 2, 3],
     [5, 6, 9],
     [5, 6, 11],
     [5, 6, 13],
     [5, 3, 13]]
df = pd.DataFrame(data=a)
df.corr()

这给了

          0         1         2
0  1.000000  0.745601  0.916579
1  0.745601  1.000000  0.544248
2  0.916579  0.544248  1.000000

def correlation_score(y_true, y_pred):
    """Scores the predictions according to the competition rules. 
    
    It is assumed that the predictions are not constant.
    
    Returns the average of each sample's Pearson correlation coefficient"""
    
    y2 = y_pred.copy()
    y2 -= y2.mean(axis=0);    y2 /= y2.std(axis=0) 
    y1 = y_true.copy(); 
    y1 -= y1.mean(axis=0);    y1 /= y1.std(axis=0) 
        
    c = (y1*y2).mean().mean()# Correlation for rescaled matrices is just matrix product and average 
        
    return c

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

下面的代码是对该定义的直接解释:

import math

def average(x):
    assert len(x) > 0
    return float(sum(x)) / len(x)

def pearson_def(x, y):
    assert len(x) == len(y)
    n = len(x)
    assert n > 0
    avg_x = average(x)
    avg_y = average(y)
    diffprod = 0
    xdiff2 = 0
    ydiff2 = 0
    for idx in range(n):
        xdiff = x[idx] - avg_x
        ydiff = y[idx] - avg_y
        diffprod += xdiff * ydiff
        xdiff2 += xdiff * xdiff
        ydiff2 += ydiff * ydiff

    return diffprod / math.sqrt(xdiff2 * ydiff2)

测试:

print pearson_def([1,2,3], [1,5,7])

返回

0.981980506062

这与Excel，这个计算器，SciPy(也是NumPy)一致，分别返回0.981980506和0.9819805060619657，和0.98198050606196574。

R:

> cor( c(1,2,3), c(1,5,7))
[1] 0.9819805

编辑:修正了一个由评论者指出的错误。