在Python中计算Pearson相关性和显着性-Java 学习之路

142

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

16 回答

你可以看一下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

回复于 2024-05-19T13:03:39+08:00

0
Pearson相关性可以用numpy的corrcoef来计算 .
```
import numpy
numpy.corrcoef(list1, list2)[0, 1]
```
回复于 2024-05-19T13:03:39+08:00
41
另一种选择可以是来自linregress的原生scipy函数，它计算：

斜率：回归线截距的斜率：回归线的截距r值：相关系数p值：假设检验的双边p值，其零假设是斜率为零stderr：标准误差估计

这是一个例子：
```
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)
```
会回报你：
```
LinregressResult(slope=0.20833333333333337, intercept=13.375, rvalue=0.14499815458068521, pvalue=0.68940144811669501, stderr=0.50261704627083648)
```
回复于 2024-05-19T13:03:39+08:00

如果你没有't feel like installing scipy, I'使用过这个快速黑客，稍微修改了Programming Collective Intelligence：

（编辑正确 . ）

from itertools import imap

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(map(lambda x: pow(x, 2), x))
  sum_y_sq = sum(map(lambda x: pow(x, 2), y))
  psum = sum(imap(lambda x, y: x * y, 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

回复于 2024-05-19T13:03:39+08:00

以下代码是the definition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])

回报

0.981980506062

这与Excel，this calculator，SciPy（也是NumPy）一致，它们分别返回0.981980506和0.9819805060619657以及0.98198050606196574 .

R：

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

EDIT ：修正了评论者指出的错误 .

回复于 2024-05-19T13:03:39+08:00

您也可以使用pandas.DataFrame.corr执行此操作：

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

回复于 2024-05-19T13:03:39+08:00

我认为我的答案应该是最容易编码和计算Pearson相关系数（PCC）的，而不是依赖于numpy / scipy .

import math

# calculates the mean
def mean(x):
    sum = 0.0
    for i in x:
         sum += i
    return sum / len(x) 

# calculates the sample standard deviation
def sampleStandardDeviation(x):
    sumv = 0.0
    for i in x:
         sumv += (i - mean(x))**2
    return math.sqrt(sumv/(len(x)-1))

# calculates the PCC using both the 2 functions above
def pearson(x,y):
    scorex = []
    scorey = []

    for i in x: 
        scorex.append((i - mean(x))/sampleStandardDeviation(x)) 

    for j in y:
        scorey.append((j - mean(y))/sampleStandardDeviation(y))

# multiplies both lists together into 1 list (hence zip) and sums the whole list   
    return (sum([i*j for i,j in zip(scorex,scorey)]))/(len(x)-1)

PCC的意义基本上是向您展示两个变量/列表是如何的.1353167_ . 值得注意的是，PCC值的范围为 from -1 to 1 . 0到1之间的值表示正相关 . 值0 =最高变化（无任何相关性） . -1到0之间的值表示负相关 .

回复于 2024-05-19T13:03:39+08:00

嗯，很多这些回复都有很长的难以阅读的代码......

在使用数组时，我建议使用numpy及其漂亮的功能：

import numpy as np
def pcc(X, Y):
   ''' Compute Pearson Correlation Coefficient. '''
   # Normalise X and Y
   X -= X.mean(0)
   Y -= Y.mean(0)
   # Standardise X and Y
   X /= X.std(0)
   Y /= Y.std(0)
   # Compute mean product
   return np.mean(X*Y)

# Using it on a random example
from random import random
X = np.array([random() for x in xrange(100)])
Y = np.array([random() for x in xrange(100)])
pcc(X, Y)

回复于 2024-05-19T13:03:39+08:00

这是基于稀疏向量的皮尔森相关的实现 . 这里的向量表示为表示为（索引，值）的元组列表 . 两个稀疏矢量可以具有不同的长度，但是在所有矢量大小上必须是相同的 . 这对于文本挖掘应用是有用的，其中矢量大小非常大，因为大多数特征是单词包，因此通常使用稀疏矢量执行计算 .

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)

回复于 2024-05-19T13:03:39+08:00

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

回复于 2024-05-19T13:03:39+08:00

这是mkh答案的变体，运行速度比它快得多，scipy.stats.pearsonr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)

回复于 2024-05-19T13:03:39+08:00

171

你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）的背景下解释你的概率 . 这是我写的一个函数来帮助它 . 它甚至可能是对的！

它基于我从http://www.vassarstats.net/rsig.html和http://en.wikipedia.org/wiki/Student%27s_t_distribution收集的信息，感谢此处发布的其他答案 .

# 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)

回复于 2024-05-19T13:03:39+08:00

16

你可以看一下这篇文章 . 这是一个详细记录的示例，用于使用pandas库（对于Python）基于来自多个文件的历史外汇货币对数据计算相关性，然后使用seaborn库生成热图图 .

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

回复于 2024-05-19T13:03:39+08:00

def pearson(x,y):
  n=len(x)
  vals=range(n)

  sumx=sum([float(x[i]) for i in vals])
  sumy=sum([float(y[i]) for i in vals])

  sumxSq=sum([x[i]**2.0 for i in vals])
  sumySq=sum([y[i]**2.0 for i in vals])

  pSum=sum([x[i]*y[i] for i in vals])
  # Calculating Pearson correlation
  num=pSum-(sumx*sumy/n)
  den=((sumxSq-pow(sumx,2)/n)*(sumySq-pow(sumy,2)/n))**.5
  if den==0: return 0
  r=num/den
  return r

回复于 2024-05-19T13:03:39+08:00

我有一个非常简单易懂的解决方案 . 对于两个相等长度的数组，Pearson系数可以很容易地计算如下：

def manual_pearson(a,b):
"""
Accepts two arrays of equal length, and computes correlation coefficient. 
Numerator is the sum of product of (a - a_avg) and (b - b_avg), 
while denominator is the product of a_std and b_std multiplied by 
length of array. 
"""
  a_avg, b_avg = np.average(a), np.average(b)
  a_stdev, b_stdev = np.std(a), np.std(b)
  n = len(a)
  denominator = a_stdev * b_stdev * n
  numerator = np.sum(np.multiply(a-a_avg, b-b_avg))
  p_coef = numerator/denominator
  return p_coef

回复于 2024-05-19T13:03:39+08:00

1
在python中使用pandas进行Pearson系数计算：我建议尝试这种方法，因为您的数据包含列表 . 您可以轻松地与数据进行交互并从控制台进行操作，因为您可以直观地显示数据结构并根据需要进行更新 . 您还可以导出数据集并将其保存并从python控制台中添加新数据以供以后分析 . 此代码更简单，包含更少的代码行 . 我假设您需要一些快速代码来筛选您的数据以进行进一步分析

例：
```
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，您没有为我发布数据，以查看数据集的大小或分析前可能需要的转换 .
回复于 2024-05-19T13:03:39+08:00

在Python中计算Pearson相关性和显着性

16 回答

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