1

当我尝试使用以下 python 代码计算马氏距离时,我在结果中得到了一些 Nan 条目。您对为什么会发生这种情况有任何见解吗?我的 data.shape = (181, 1500)

from scipy.spatial.distance import pdist, squareform

data_log = log2(data + 1) # A log transform that I usually apply to my data
data_centered = data_log - data_log.mean(0) # zero centering
D = squareform( pdist(data_centered, 'mahalanobis' ) )

我也试过:

data_standard = data_centered / data_centered.std(0, ddof=1)
D = squareform( pdist(data_standard, 'mahalanobis' ) )

还得到了nans。输入没有损坏,并且可以很好地计算其他距离,例如相关距离。出于某种原因,当我减少功能数量时,我不再获得 Nans。例如,以下示例没有得到任何 Nan:

D = squareform( pdist(data_centered[:,:200], 'mahalanobis' ) )
D = squareform( pdist(data_centered[:,180:480], 'mahalanobis' ) )

而其他人则得到Nans:

D = squareform( pdist(data_centered[:,:300], 'mahalanobis' ) )
D = squareform( pdist(data_centered[:,180:600], 'mahalanobis' ) )

有什么线索吗?如果输入的某些条件不满足,这是预期的行为吗?

4

1 回答 1

4

您的观测值少于特征,因此Vscipy 代码计算的协方差矩阵是奇异的。代码不检查这一点,并盲目地计算协方差矩阵的“逆”。因为这个数值计算的逆基本上是垃圾,所以乘积(x-y)*inv(V)*(x-y)(其中xy是观察值)可能会变成负数。然后该值的平方根导致nan

例如,这个数组也会产生一个nan

In [265]: x
Out[265]: 
array([[-1. ,  0.5,  1. ,  2. ,  2. ],
       [ 2. ,  1. ,  2.5, -1.5,  1. ],
       [ 1.5, -0.5,  1. ,  2. ,  2.5]])

In [266]: squareform(pdist(x, 'mahalanobis'))
Out[266]: 
array([[ 0.        ,         nan,  1.90394328],
       [        nan,  0.        ,         nan],
       [ 1.90394328,         nan,  0.        ]])

这是“手动”完成的马氏计算:

In [279]: V = np.cov(x.T)

理论上V是单数;以下值实际上为 0:

In [280]: np.linalg.det(V)
Out[280]: -2.968550671342364e-47

inv没有看到问题,并返回一个逆:

In [281]: VI = np.linalg.inv(V)

让我们计算 和 之间的距离x[0],并验证当我们使用 时x[2]我们得到了相同的非 nan 值(1.9039):pdistVI

In [295]: delta = x[0] - x[2]

In [296]: np.dot(np.dot(delta, VI), delta)
Out[296]: 3.625

In [297]: np.sqrt(np.dot(np.dot(delta, VI), delta))
Out[297]: 1.9039432764659772

当我们尝试计算 和 之间的距离时,会发生以下x[0]情况x[1]

In [300]: delta = x[0] - x[1]

In [301]: np.dot(np.dot(delta, VI), delta)
Out[301]: -1.75

然后该值的平方根给出nan


在 scipy 0.16(将于 2015 年 6 月发布)中,您将收到错误而不是 nan 或垃圾。错误消息描述了问题:

In [4]: x = array([[-1. ,  0.5,  1. ,  2. ,  2. ],
   ...:        [ 2. ,  1. ,  2.5, -1.5,  1. ],
   ...:        [ 1.5, -0.5,  1. ,  2. ,  2.5]])

In [5]: pdist(x, 'mahalanobis')
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-5-a3453ff6fe48> in <module>()
----> 1 pdist(x, 'mahalanobis')

/Users/warren/local_scipy/lib/python2.7/site-packages/scipy/spatial/distance.pyc in pdist(X, metric, p, w, V, VI)
   1298                                      "singular. For observations with %d "
   1299                                      "dimensions, at least %d observations "
-> 1300                                      "are required." % (m, n, n + 1))
   1301                 V = np.atleast_2d(np.cov(X.T))
   1302                 VI = _convert_to_double(np.linalg.inv(V).T.copy())

ValueError: The number of observations (3) is too small; the covariance matrix is singular. For observations with 5 dimensions, at least 6 observations are required.
于 2015-04-18T14:01:31.527 回答