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

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/

其他回答

本文给出了一种基于稀疏向量的pearson相关的实现方法。这里的向量表示为(index, value)表示的元组列表。两个稀疏向量可以是不同的长度,但总的向量大小必须是相同的。这对于文本挖掘应用程序非常有用,其中向量大小非常大,因为大多数特征都是单词包,因此通常使用稀疏向量执行计算。

def get_pearson_corelation(self, first_feature_vector=[], second_feature_vector=[], length_of_featureset=0):
    indexed_feature_dict = {}
    if first_feature_vector == [] or second_feature_vector == [] or length_of_featureset == 0:
        raise ValueError("Empty feature vectors or zero length of featureset in get_pearson_corelation")

    sum_a = sum(value for index, value in first_feature_vector)
    sum_b = sum(value for index, value in second_feature_vector)

    avg_a = float(sum_a) / length_of_featureset
    avg_b = float(sum_b) / length_of_featureset

    mean_sq_error_a = sqrt((sum((value - avg_a) ** 2 for index, value in first_feature_vector)) + ((
        length_of_featureset - len(first_feature_vector)) * ((0 - avg_a) ** 2)))
    mean_sq_error_b = sqrt((sum((value - avg_b) ** 2 for index, value in second_feature_vector)) + ((
        length_of_featureset - len(second_feature_vector)) * ((0 - avg_b) ** 2)))

    covariance_a_b = 0

    #calculate covariance for the sparse vectors
    for tuple in first_feature_vector:
        if len(tuple) != 2:
            raise ValueError("Invalid feature frequency tuple in featureVector: %s") % (tuple,)
        indexed_feature_dict[tuple[0]] = tuple[1]
    count_of_features = 0
    for tuple in second_feature_vector:
        count_of_features += 1
        if len(tuple) != 2:
            raise ValueError("Invalid feature frequency tuple in featureVector: %s") % (tuple,)
        if tuple[0] in indexed_feature_dict:
            covariance_a_b += ((indexed_feature_dict[tuple[0]] - avg_a) * (tuple[1] - avg_b))
            del (indexed_feature_dict[tuple[0]])
        else:
            covariance_a_b += (0 - avg_a) * (tuple[1] - avg_b)

    for index in indexed_feature_dict:
        count_of_features += 1
        covariance_a_b += (indexed_feature_dict[index] - avg_a) * (0 - avg_b)

    #adjust covariance with rest of vector with 0 value
    covariance_a_b += (length_of_featureset - count_of_features) * -avg_a * -avg_b

    if mean_sq_error_a == 0 or mean_sq_error_b == 0:
        return -1
    else:
        return float(covariance_a_b) / (mean_sq_error_a * mean_sq_error_b)

单元测试:

def test_get_get_pearson_corelation(self):
    vector_a = [(1, 1), (2, 2), (3, 3)]
    vector_b = [(1, 1), (2, 5), (3, 7)]
    self.assertAlmostEquals(self.sim_calculator.get_pearson_corelation(vector_a, vector_b, 3), 0.981980506062, 3, None, None)

    vector_a = [(1, 1), (2, 2), (3, 3)]
    vector_b = [(1, 1), (2, 5), (3, 7), (4, 14)]
    self.assertAlmostEquals(self.sim_calculator.get_pearson_corelation(vector_a, vector_b, 5), -0.0137089240555, 3, None, None)
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

从Python 3.10开始,Pearson的相关系数(statistics.correlation)可以直接在标准库中获得:

from statistics import correlation

# a = [15, 12, 8, 8, 7, 7, 7, 6, 5, 3]
# b = [10, 25, 17, 11, 13, 17, 20, 13, 9, 15]
correlation(a, b)
# 0.1449981545806852

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