两个特征矩阵的高效成对相关 [英] Efficient pairwise correlation for two matrices of features
问题描述
在Python中，我需要找到矩阵A中的所有特征与矩阵B中的所有特征之间的成对相关性.特别是，我很感兴趣地发现A中给定特征在B中的所有特征之间具有最强的Pearson相关性.我不在乎最强的相关性是正还是负.
In Python I need to find the pairwise correlation between all features in a matrix A and all features in a matrix B. In particular, I am interesting in finding the strongest Pearson correlation that a given feature in A has across all features in B. I do not care whether the strongest correlation is positive or negative.
我使用下面的两个循环和scipy进行了效率低下的实现.但是，我想使用np.corrcoef或其他类似方法来有效地计算它.矩阵A的形状为40000x400，B的形状为40000x1440.我为有效执行此操作所做的尝试可以在下面看到为方法find_max_absolute_corr(A,B).但是，它失败并显示以下错误:
I've done a inefficient implementation using two loops and scipy below. However, I'd like to use np.corrcoef or another similar method to compute it efficiently. Matrix A has shape 40000x400 and B has shape 40000x1440. My attempt at doing it efficiently can be seen below as the method find_max_absolute_corr(A,B). However, it fails with the following error:
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]
推荐答案
基于该公式，您可以轻松地向量化，因为A和B的列的成对计算彼此独立.这是使用 broadcasting -
Based on that formula, you can vectorized easily as the pairwise computations of columns from A and B are independent of each other. Here's one vectorized solution using broadcasting -
# 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])]
样品运行-
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-
2) Original loopy code run -
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)建议的代码运行-
3) Proposed code run -
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
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