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用Python计算Pearson相关性和意义 (10)

Pearson相关性可以用numpy的corrcoef来计算。

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

我正在寻找一个以输入两个列表为参数的函数,并返回Pearson相关性相关性的显着性。


以下代码是对定义的直接解释:

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])

回报

0.981980506062

这与Excel, 这个计算器SciPy (也是corrcoef )一致,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 :修正了评论者指出的错误。



你可能想知道如何在寻找特定方向上的相关性(负相关或正相关)的情况下解释你的概率。下面是我写的一个函数来帮助解决这个问题。 它甚至可能是正确的!

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)

嗯,其中很多答复都很长且很难读取代码...

在使用数组时,我建议使用numpy来处理漂亮的功能:

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)

如果你不喜欢安装scipy,我已经使用了这个快速入门,稍微修改了编程集体智慧

(为正确性编辑。)

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

这是一个基于稀疏向量的皮尔森相关实现。 这里的向量表示为以(索引,值)表示的元组列表。 两个稀疏矢量可以具有不同的长度,但是所有矢量大小必须相同。 这对于文本挖掘应用非常有用,因为大多数特征都是单词包,因此矢量大小非常大,因此计算通常使用稀疏矢量执行。

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)

这是使用numpy的皮尔逊相关函数的实现:


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


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

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






scipy