在Python中计算Pearson相关性和重要性


Answers:


202

您可以看一下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

2
两个字典的相关系数如何?
user702846 2013年

2
@ user702846在2xN矩阵上定义了Pearson相关性。没有普遍适用的方法将两个字典转换为2xN矩阵,但是您可以使用与字典键的交点对应的字典值对数组。
winerd


56

一种替代可以是来自天然SciPy的功能linregress其计算:

斜率:回归线的斜率

截距:回归线的截距

r值:相关系数

p值:假设检验的两侧p值,其零假设是斜率为零

stderr:估算的标准误

这是一个示例:

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)

将返回您:

LinregressResult(slope=0.20833333333333337, intercept=13.375, rvalue=0.14499815458068521, pvalue=0.68940144811669501, stderr=0.50261704627083648)

2
好的答案-迄今为止最有用的信息。也适用于两排pandas.DataFrame:lineregress(two_row_df)
dmeu

辉煌的答案。如果您想到的话,也非常直观
Raghuram,

37

如果您不想安装scipy,请使用此快速技巧,该技巧已从Programming Collective Intelligence进行了稍微修改

(为确保正确性进行了编辑。)

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

2
我惊讶地发现这与Excel,NumPy和R不同。请参阅stackoverflow.com/questions/3949226/…
dfrankow 2011年

2
正如另一位评论者指出的那样,这有一个float / int错误。我认为sum_y / n是整数的整数除法。如果使用sum_x = float(sum(x))和sum_y = float(sum(y)),则可以使用。
dfrankow 2011年

@dfrankow我认为这是因为imap无法处理浮动。蟒蛇给出了一个TypeError: unsupported operand type(s) for -: 'itertools.imap' and 'float'num = psum - (sum_x * sum_y/n)
alvas

4
作为样式说明,Python反对这种不必要的map使用(支持列表理解)
Maxim Khesin

14
就像评论一样,认为scipy等人的库是由知道很多数值分析的人开发的。这可以避免您遇到许多常见的陷阱(例如,X或Y中的数字过大或过少都会导致灾难性的抵消)
geekazoid 2014年

32

以下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])

退货

0.981980506062

这符合Excel中,这个计算器SciPy的(也NumPy的),它分别返回0.981980506和0.9819805060619657和0.98198050606196574。

R

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

编辑:修复了评论者指出的错误。


4
当心变量的类型!您遇到了int / float问题。在sum(x) / len(x)您分割整数,而不是浮点数。因此sum([1,5,7]) / len([1,5,7]) = 13 / 3 = 4,根据整数除法(而您要13. / 3. = 4.33...)。要解决此问题,请将此行重写为float(sum(x)) / float(len(x))(一个浮点数就足够了,因为Python自动将其转换)。
Piotr Migdal

您的代码不适用于以下情况:[10,10,10],[0,0,0]或[10,10],[10,0]。甚至[10,10],[10,10]
madCode 2012年

4
在任何情况下都没有定义相关系数。将它们放入R会为所有三个返回“ NA”。
dfrankow 2012年

28

您也可以使用进行此操作pandas.DataFrame.corr

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

          0         1         2
0  1.000000  0.745601  0.916579
1  0.745601  1.000000  0.544248
2  0.916579  0.544248  1.000000

5
这只是相关性而没有意义
Ivelin

12

我认为我的答案应该最简单地编码和理解计算Pearson相关系数(PCC)的步骤,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显示两个变量/列表之间的相关程度如何。重要的是要注意PCC值的范围是-1至1。0到1之间的值表示正相关。值0 =最大变化(无相关性)。-1至0之间的值表示负相关。


2
请注意,Python具有内置sum函数。
bfontaine 2015年

5
它在2个具有500多个值的列表上具有惊人的复杂性和较慢的性能。
Nikolay Fominyh,2016年

9

使用python中的pandas进行Pearson系数计算:由于您的数据包含列表,建议您尝试这种方法。与数据进行交互并从控制台对其进行操作将很容易,因为您可以可视化数据结构并根据需要进行更新。您还可以导出数据集并保存它,并从python控制台中添加新数据以供以后分析。该代码更简单,包含更少的代码行。我假设您需要一些快速的代码行来筛选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之前可能需要进行的转换。


您好,欢迎来到StackOverflow!尝试在回答开始时添加简短说明,说明为什么选择此代码以及在这种情况下如何应用该代码!
特里斯托

8

嗯,这些响应中的许多响应都有很长且很难阅读的代码...

我建议在使用数组时使用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)

尽管我非常喜欢这个答案,但我还是建议在函数内部复制/克隆X和Y。否则,两者都会被更改,这可能不是期望的行为。
antonimmo

7

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


7

这是mkh答案的一种变体,它比使用numba和scipy.stats.pearsonr运行得快得多。

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)

5

这是基于稀疏向量的皮尔逊相关性的实现。这里的向量表示为表示为(索引,值)的元组列表。两个稀疏向量的长度可以不同,但​​在所有向量上,大小必须相同。这对于文本挖掘应用很有用,因为大多数特征都是单词包,因此向量大小非常大,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)

3

我有一个非常简单易懂的解决方案。对于长度相等的两个数组,皮尔逊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

1

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)的情况下解释您的概率。这是我编写的用于帮助实现此功能的函数。甚至可能是对的!

它基于我从http://www.vassarstats.net/rsig.htmlhttp://en.wikipedia.org/wiki/Student%27s_t_distribution收集的信息,这要归功于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)


0
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

仅代码答案不被视为良好实践。请考虑添加几句话来解释您的代码如何解决该问题。(阅读有关如何回答SO问题的帮助页面)
Yannis
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