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

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)

您可能想知道如何在寻找特定方向的相关性(负相关或正相关)的上下文中解释您的概率。这是我写的一个函数。它甚至可能是正确的!

这是基于我从http://www.vassarstats.net/rsig.html和http://en.wikipedia.org/wiki/Student%27s_t_distribution上收集到的信息，感谢这里发布的其他答案。

# Given (possibly random) variables, X and Y, and a correlation direction,
# returns:
#  (r, p),
# where r is the Pearson correlation coefficient, and p is the probability
# that there is no correlation in the given direction.
#
# direction:
#  if positive, p is the probability that there is no positive correlation in
#    the population sampled by X and Y
#  if negative, p is the probability that there is no negative correlation
#  if 0, p is the probability that there is no correlation in either direction
def probabilityNotCorrelated(X, Y, direction=0):
    x = len(X)
    if x != len(Y):
        raise ValueError("variables not same len: " + str(x) + ", and " + \
                         str(len(Y)))
    if x < 6:
        raise ValueError("must have at least 6 samples, but have " + str(x))
    (corr, prb_2_tail) = stats.pearsonr(X, Y)

    if not direction:
        return (corr, prb_2_tail)

    prb_1_tail = prb_2_tail / 2
    if corr * direction > 0:
        return (corr, prb_1_tail)

    return (corr, 1 - prb_1_tail)

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

你可以看看这篇文章。这是一个使用pandas库(适用于Python)根据多个文件的历史外汇货币对数据计算相关性的示例，然后使用seaborn库生成热图图。

http://www.tradinggeeks.net/2015/08/calculating-correlation-in-python/

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)