numpy中的加权协方差矩阵 [英] weighted covariance matrix in numpy

问题描述

我想计算p个数量的n个测量值的协方差C，其中每个单独的数量测量值都有自己的权重.也就是说，我的权重数组W与我的数量数组Q具有相同的形状(n除以p).原生np.cov()函数仅支持赋予各个度量的权重(即长度为n的向量).

I want to compute the covariance C of n measurements of p quantities, where each individual quantity measurement is given its own weight. That is, my weight array W has the same shape as my quantity array Q (n by p). The native np.cov() function only supports weights given to individual measurements (i.e., a vector of length n).

我可以通过p矩阵初始化p并进行迭代，但是如果p很大，那么这将是一个非常缓慢的过程.

I can initialize a p by p matrix and iterate, but if p is large, then it's a very slow process.

由于已知Q对于每个量(Q的列)均值为零，所以C的每个元素的显式公式为

Since Q is known to have mean zero for each quantity (column of Q), the explicit formula for each element of C is

C[i,j] = np.sum(
    Q[:, i] * Q[:, j] * W[:, i] * W[:, j]) / np.sum(W[:, i] * W[:, j])

如果我将分子重新设置为Q[:, i] * W[:, i] * Q[:, j] * W[:, j]，看来我应该能够对Q * W的列进行乘法和求和，然后类似地进行分母(使用W * W除外).

If I rearrange the numerator to be Q[:, i] * W[:, i] * Q[:, j] * W[:, j], it seems like I should be able to multiply and sum columns of Q * W, and then do the denominator similarly (except using W * W).

有没有办法用np.einsum()做到这一点?

Is there a way to do this with np.einsum()?

对于测试，让我们定义以下内容:

For testing, let's define the following:

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]])

推荐答案

您可以将非常有效的矩阵乘法与

You can use the very efficient matrix-multiplication with np.dot -

QW = Q*W
C = QW.T.dot(QW)/W.T.dot(W)

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