我将高斯核密度估计器拟合到一个变量,该变量是两个向量的差,称为“diff”,如下所示: gaussian_kde_covfact(diff, smoothing_param) - 其中 gaussian_kde_covfact 定义为:
class gaussian_kde_covfact(stats.gaussian_kde):
def __init__(self, dataset, covfact = 'scotts'):
self.covfact = covfact
scipy.stats.gaussian_kde.__init__(self, dataset)
def _compute_covariance_(self):
'''not used'''
self.inv_cov = np.linalg.inv(self.covariance)
self._norm_factor = sqrt(np.linalg.det(2*np.pi*self.covariance)) * self.n
def covariance_factor(self):
if self.covfact in ['sc', 'scotts']:
return self.scotts_factor()
if self.covfact in ['si', 'silverman']:
return self.silverman_factor()
elif self.covfact:
return float(self.covfact)
else:
raise ValueError, \
'covariance factor has to be scotts, silverman or a number'
def reset_covfact(self, covfact):
self.covfact = covfact
self.covariance_factor()
self._compute_covariance()
这可行,但有一个极端情况,即 diff 是全 0 的向量。在这种情况下,我收到错误:
File "/srv/pkg/python/python-packages/python26/scipy/scipy-0.7.1/lib/python2.6/site-packages/scipy/stats/kde.py", line 334, in _compute_covariance
self.inv_cov = linalg.inv(self.covariance)
File "/srv/pkg/python/python-packages/python26/scipy/scipy-0.7.1/lib/python2.6/site-packages/scipy/linalg/basic.py", line 382, in inv
if info>0: raise LinAlgError, "singular matrix"
numpy.linalg.linalg.LinAlgError: singular matrix
有什么办法可以解决这个问题?在这种情况下,我希望它返回一个密度,该密度基本上完全在 0 的差异处达到峰值,其他地方没有质量。
谢谢。