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

Pearson相关性可以用numpy的corrcoef来计算。

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

其他回答

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

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)
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

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。