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我一直在使用 rpy2 来计算测试向量和先验分布之间的马氏距离。我想放弃 rpy2 并转向 scipy，但是当我测试它时，rpy2 和 scipy 不会返回相同的结果。这是我的示例代码。

import numpy as np
from scipy import linalg
from scipy.spatial.distance import mahalanobis as mahalanobis
import rpy2.robjects as robjects

# The vector to test.
test_values = [692.5816522801106, 1421.4737901031651, 6.117859, 7.259449]
test_values_r = robjects.FloatVector(test_values)
test_values_np = np.array(test_values)

# The covariance matrix from the prior distribution
covs = [15762.87, 13486.23, 34.61164, 22.15451, 
        13486.23, 36003.67, 33.8431, 30.52712, 
        34.61164, 33.8431, 0.4143354, 0.1125765, 
        22.15451, 30.52712, 0.1125765, 0.2592451]
covs_np = np.reshape(np.array(covs), (4,-1))
covs_r  = robjects.r["matrix"](robjects.FloatVector(covs), nrow = 4)

# The means of the prior distribution
centers = [808.0645, 1449.711, 4.8443, 4.95776]
centers_np = np.array(centers)
centers_r  = robjects.FloatVector(centers)

r_dist = robjects.r["mahalanobis"](test_values_r, centers_r, covs_r)
# <FloatVector - Python:0x1052275a8 / R:0x10701bfa8>
# [29.782287]

np_dist = mahalanobis(test_values_np, centers_np, linalg.inv(covs_np))
# 5.4573150053873185

我错过了一些明显的东西吗？

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1 回答 1

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该R函数返回平方的Mahalanobis 距离（例如，请参见此处）。

因此：

>>> r_dist[0]
29.782287068025585
>>> np_dist
5.4573150053873185
>>> np_dist**2 - r_dist[0]
3.5527136788005009e-15
于 2013-10-31T13:29:40.013 回答