如何计算一个非方阵的Cholesky分解,以便用'numpy'计算马氏距离?

vc6uscn9  于 2022-12-04  发布在  其他
关注(0)|答案(1)|浏览(179)

如何计算一个非方阵的Cholesky分解,以计算马氏距离numpy

def get_fitting_function(G):
    print(G.shape) #(14L, 11L) --> 14 samples of dimension 11
    g_mu = G.mean(axis=0) 
    #Cholesky decomposition uses half of the operations as LU
    #and is numerically more stable.
    L = np.linalg.cholesky(G)

    def fitting_function(g):
        x = g - g_mu
        z = np.linalg.solve(L, x)
        #Mahalanobis Distance 
        MD = z.T*z
        return math.sqrt(MD)
    return fitting_function  

C:\Users\Matthias\CV\src\fitting_function.py in get_fitting_function(G)
     22     #Cholesky decomposition uses half of the operations as LU
     23     #and is numerically more stable.
---> 24     L = np.linalg.cholesky(G)
     25 
     26     def fitting_function(g):

C:\Users\Matthias\AppData\Local\Enthought\Canopy\User\lib\site-packages\numpy\linalg\linalg.pyc in cholesky(a)
    598     a, wrap = _makearray(a)
    599     _assertRankAtLeast2(a)
--> 600     _assertNdSquareness(a)
    601     t, result_t = _commonType(a)
    602     signature = 'D->D' if isComplexType(t) else 'd->d'

C:\Users\Matthias\AppData\Local\Enthought\Canopy\User\lib\site-packages\numpy\linalg\linalg.pyc in _assertNdSquareness(*arrays)
    210     for a in arrays:
    211         if max(a.shape[-2:]) != min(a.shape[-2:]):
--> 212             raise LinAlgError('Last 2 dimensions of the array must be square')
    213 
    214 def _assertFinite(*arrays):

LinAlgError: Last 2 dimensions of the array must be square 

    LinAlgError: Last 2 dimensions of the array must be square

基于Matlab实现:马氏距离协方差矩阵求逆
编辑:chol(a)=linalg.cholesky(a).T矩阵的Cholesky分解(matlab中的chol(a)返回上三角矩阵,但linalg.cholesky(a)返回下三角矩阵)(来源:(第10页)
编辑2:

G -= G.mean(axis=0)[None, :]
C = (np.dot(G, G.T) / float(G.shape[0]))
#Cholesky decomposition uses half of the operations as LU
#and is numerically more stable.
L = np.linalg.cholesky(C).T

因此,如果D=x^t.S^-1.x=x^t.(L.L^t)^-1.x=x^t. L^t.x=z^t.z

35g0bw71

35g0bw711#

我不相信你能做到。Cholesky分解不仅需要一个方阵,还需要一个Hermitian矩阵和一个正定矩阵来保证唯一性。它基本上是一个LU分解,条件是L = U '。事实上,该算法经常被用作数值检验给定矩阵是否正定的方法。参见Wikipedia
也就是说,根据定义,协方差矩阵是对称半正定的,所以你应该能够对它做cholesky运算。
编辑:当你计算它的时候,你的矩阵C=np.dot(G, G.T)应该是对称的,但是可能有什么地方不对。你可以尝试强制它对称C = ( C + C.T) /2.0,然后再试chol(C)

相关问题