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)

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

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

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

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

如果你不喜欢安装scipy，我使用了这个快速的hack，稍微修改了Programming Collective Intelligence:

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(xi*xi for xi in x)
  sum_y_sq = sum(yi*yi for yi in y)
  psum = sum(xi*yi for xi, yi in zip(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