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
其他回答
对此,我有一个非常简单易懂的解决方案。对于两个长度相等的数组,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
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)
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)的上下文中解释您的概率。这是我写的一个函数。它甚至可能是正确的!
这是基于我从http://www.vassarstats.net/rsig.html和http://en.wikipedia.org/wiki/Student%27s_t_distribution上收集到的信息,感谢这里发布的其他答案。
# 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)
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)
嗯,很多回复的代码都很长,很难读…
我建议在处理数组时使用numpy及其漂亮的特性:
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)
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