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
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)
其他回答
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 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
与其依赖numpy/scipy,我认为我的答案应该是最容易编码和理解计算Pearson相关系数(PCC)的步骤。
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的意义基本上是向你展示两个变量/列表的相关性有多强。 需要注意的是,PCC值的范围是-1到1。 0到1之间的值表示正相关。 0值=最高变异(没有任何相关性)。 -1到0之间的值表示负相关。
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)
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
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