我想计算协方差C
of n
的测量p
数量,其中每个单独的数量测量都有自己的权重。也就是我的权重数组W
与我的数量数组具有相同的形状Q
(n
by p
)。当地人np.cov()
函数仅支持赋予各个测量值的权重(即长度向量n
).
我可以初始化一个p
by p
矩阵并迭代,但如果p
很大,那么这是一个非常慢的过程。
Since Q
已知每个数量的均值为零(Q
),每个元素的显式公式C
is
C[i,j] = np.sum(
Q[:, i] * Q[:, j] * W[:, i] * W[:, j]) / np.sum(W[:, i] * W[:, j])
如果我将分子重新排列为Q[:, i] * W[:, i] * Q[:, j] * W[:, j]
,看来我应该能够对以下列进行乘法和求和Q * W
,然后类似地计算分母(除了使用W * W
).
有没有办法做到这一点np.einsum()
?
为了进行测试,我们定义以下内容:
C = array([[ 1. , 0.1 , 0.2 ], # set this beforehand, to test whether
[ 0.1 , 0.5 , 0.15], # we get the correct result
[ 0.2 , 0.15, 0.75]])
Q = array([[-0.6084634 , 0.16656143, -1.04490324],
[-1.51164337, -0.96403094, -2.37051952],
[-0.32781346, -0.19616374, -1.32591578],
[-0.88371729, 0.20877833, -0.52074272],
[-0.67987913, -0.84458226, 0.02897935],
[-2.01924756, -0.51877396, -0.68483981],
[ 1.64600477, 0.67620595, 1.24559591],
[ 0.82554885, 0.14884613, -0.15211434],
[-0.88119527, 0.11663335, -0.31522598],
[-0.14830668, 1.26906561, -0.49686309]])
W = array([[ 1.01133857, 0.91962164, 1.01897898],
[ 1.09467975, 0.91191381, 0.90150961],
[ 0.96334661, 1.00759046, 1.01638749],
[ 1.04827001, 0.95861001, 1.01248969],
[ 0.91572506, 1.09388218, 1.03616461],
[ 0.9418178 , 1.07210878, 0.90431879],
[ 1.0093642 , 1.00408472, 1.07570172],
[ 0.92203074, 1.00022631, 1.09705542],
[ 0.99775598, 0.01000000, 0.94996408],
[ 1.02996389, 1.01224303, 1.00331465]])