我正在寻找一个函数,它将两个列表作为输入,并返回Pearson相关性,以及相关性的重要性。


当前回答

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

其他回答

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

Pearson相关性可以用numpy的corrcoef来计算。

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

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