两个特征矩阵的有效成对相关

Question

在 Python 中，我需要找到矩阵 A 中所有特征与矩阵 B 中所有特征之间的成对相关性。特别是，我感兴趣的是找到 A 中给定特征与 B 中所有特征的最强 Pearson 相关性。我不关心最强相关性是正相关还是负相关。

我使用下面的两个循环和 scipy 完成了一个低效的实现。但是，我想使用 np.corrcoef 或其他类似的方法来有效地计算它。矩阵 A 的形状为 40000x400，B 的形状为 40000x1440。我尝试有效地做到这一点可以在下面看到作为方法 find_max_absolute_corr(A,B)。但是，它失败并出现以下错误：

ValueError: all the input array dimensions except for the concatenation axis must match exactly.

import numpy as np
from scipy.stats import pearsonr


def find_max_absolute_corr(A, B):
    """ Finds for each feature in `A` the highest Pearson
        correlation across all features in `B`. """

    max_corr_A = np.zeros((A.shape[1]))    

    for A_col in range(A.shape[1]):
        print "Calculating {}/{}.".format(A_col+1, A.shape[1])

        metric = A[:,A_col]
        pearson = np.corrcoef(B, metric, rowvar=0)

        # takes negative correlations into account as well
        min_p = min(pearson)
        max_p = max(pearson)
        max_corr_A[A_col] = max_absolute(min_p, max_p)

    return max_corr_A


def max_absolute(min_p, max_p):
    if np.isnan(min_p) or np.isnan(max_p):
        raise ValueError("NaN correlation.")
    if abs(max_p) > abs(min_p):
        return max_p
    else:
        return min_p


if __name__ == '__main__':

    A = np.array(
        [[10, 8.04, 9.14, 7.46],
         [8, 6.95, 8.14, 6.77],
         [13, 7.58, 8.74, 12.74],
         [9, 8.81, 8.77, 7.11],
         [11, 8.33, 9.26, 7.81]])

    B = np.array(
        [[-14, -9.96, 8.10, 8.84, 8, 7.04], 
         [-6, -7.24, 6.13, 6.08, 5, 5.25], 
         [-4, -4.26, 3.10, 5.39, 8, 5.56], 
         [-12, -10.84, 9.13, 8.15, 5, 7.91], 
         [-7, -4.82, 7.26, 6.42, 8, 6.89]])

    # simple, inefficient method
    for A_col in range(A.shape[1]): 
        high_corr = 0
        for B_col in range(B.shape[1]):
            corr,_ = pearsonr(A[:,A_col], B[:,B_col])
            high_corr = max_absolute(high_corr, corr)
        print high_corr

    # -0.161314601631
    # 0.956781516149
    # 0.621071009239
    # -0.421539304112        

    # efficient method
    max_corr_A = find_max_absolute_corr(A, B)
    print max_corr_A

    # [-0.161314601631,
    # 0.956781516149,
    # 0.621071009239,
    # -0.421539304112]

Answer 1

似乎 scipy.stats.pearsonr 遵循应用于 A & B -

的逐列对的 Pearson 相关系数公式的定义

基于该公式，您可以轻松矢量化，因为 A 和 B 的列的成对计算彼此独立。这是一个使用 broadcasting -

的矢量化解决方案

# Get number of rows in either A or B
N = B.shape[0]

# Store columnw-wise in A and B, as they would be used at few places
sA = A.sum(0)
sB = B.sum(0)

# Basically there are four parts in the formula. We would compute them one-by-one
p1 = N*np.einsum('ij,ik->kj',A,B)
p2 = sA*sB[:,None]
p3 = N*((B**2).sum(0)) - (sB**2)
p4 = N*((A**2).sum(0)) - (sA**2)

# Finally compute Pearson Correlation Coefficient as 2D array 
pcorr = ((p1 - p2)/np.sqrt(p4*p3[:,None]))

# Get the element corresponding to absolute argmax along the columns 
out = pcorr[np.nanargmax(np.abs(pcorr),axis=0),np.arange(pcorr.shape[1])]

样本运行-

1) 输入：

In [12]: A
Out[12]: 
array([[ 10.  ,   8.04,   9.14,   7.46],
       [  8.  ,   6.95,   8.14,   6.77],
       [ 13.  ,   7.58,   8.74,  12.74],
       [  9.  ,   8.81,   8.77,   7.11],
       [ 11.  ,   8.33,   9.26,   7.81]])

In [13]: B
Out[13]: 
array([[-14.  ,  -9.96,   8.1 ,   8.84,   8.  ,   7.04],
       [ -6.  ,  -7.24,   6.13,   6.08,   5.  ,   5.25],
       [ -4.  ,  -4.26,   3.1 ,   5.39,   8.  ,   5.56],
       [-12.  , -10.84,   9.13,   8.15,   5.  ,   7.91],
       [ -7.  ,  -4.82,   7.26,   6.42,   8.  ,   6.89]])

2) 原始循环代码运行 -

In [14]: high_corr_out = np.zeros(A.shape[1])
    ...: for A_col in range(A.shape[1]): 
    ...:     high_corr = 0
    ...:     for B_col in range(B.shape[1]):
    ...:         corr,_ = pearsonr(A[:,A_col], B[:,B_col])
    ...:         high_corr = max_absolute(high_corr, corr)
    ...:     high_corr_out[A_col] = high_corr
    ...:     

In [15]: high_corr_out
Out[15]: array([ 0.8067843 ,  0.95678152,  0.74016181, -0.85127779])

3）建议代码运行-

In [16]: N = B.shape[0]
    ...: sA = A.sum(0)
    ...: sB = B.sum(0)
    ...: p1 = N*np.einsum('ij,ik->kj',A,B)
    ...: p2 = sA*sB[:,None]
    ...: p3 = N*((B**2).sum(0)) - (sB**2)
    ...: p4 = N*((A**2).sum(0)) - (sA**2)
    ...: pcorr = ((p1 - p2)/np.sqrt(p4*p3[:,None]))
    ...: out = pcorr[np.nanargmax(np.abs(pcorr),axis=0),np.arange(pcorr.shape[1])]
    ...: 

In [17]: pcorr # Pearson Correlation Coefficient array
Out[17]: 
array([[ 0.41895565, -0.5910935 , -0.40465987,  0.5818286 ],
       [ 0.66609445, -0.41950457,  0.02450215,  0.64028344],
       [-0.64953314,  0.65669916,  0.30836196, -0.85127779],
       [-0.41917583,  0.59043266,  0.40364532, -0.58144102],
       [ 0.8067843 ,  0.07947386,  0.74016181,  0.53165395],
       [-0.1613146 ,  0.95678152,  0.62107101, -0.4215393 ]])

In [18]: out # elements corresponding to absolute argmax along columns
Out[18]: array([ 0.8067843 ,  0.95678152,  0.74016181, -0.85127779])

运行时测试 -

In [36]: A = np.random.rand(4000,40)

In [37]: B = np.random.rand(4000,144)

In [38]: np.allclose(org_app(A,B),proposed_app(A,B))
Out[38]: True

In [39]: %timeit org_app(A,B) # Original approach
1 loops, best of 3: 1.35 s per loop

In [40]: %timeit proposed_app(A,B) # Proposed vectorized approach
10 loops, best of 3: 39.1 ms per loop

Answer 2

根据个人经验添加上述答案，

p1 = N*np.dot(B.T,A)

与

相比，我的工作速度更快

p1 = N*np.einsum('ij,ik->kj',A,B)

当 A 和 B 是大维矩阵时尤其如此。

两个特征矩阵的有效成对相关

Efficient pairwise correlation for two matrices of features

python

numpy

matrix

vectorization

correlation