如何在协方差矩阵中找到退化的行/列

本文介绍了如何在协方差矩阵中找到退化的行/列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在使用 numpy.cov 从超过 400 个时间序列的数据集创建协方差矩阵.使用 linalg.det 给我一个零值，所以矩阵是奇异的.我可以使用 linalg.svd 来查看秩比列数少 2，因此在协方差矩阵中的某处我有一些线性组合来使矩阵退化.我在底层时间序列上使用了 corrcoef，但没有相关性 > 0.78，所以在那里不明显.有人可以建议一种方法来确定退化列的位置.谢谢.

I am using numpy.cov to create a covariance matrix from a dataset of over 400 time series. Using linalg.det gives me a value of zero so matrix is singular. I can use linalg.svd to see that the rank is two less than the number of columns so somewhere in the covariance matrix I have some linear combinations to make the matrix degenerate. I have used corrcoef on the underlying timeseries but no correlation > 0.78 so not obvious there. Can someone suggest a method to determine the location of the degenerate columns. Thank you.

推荐答案

如果对矩阵A进行QR分解，R 沿对角线的非零值对应于 A 的线性独立列.

If you take the QR decomposition of a matrix A, the columns of R with a non-zero value along the diagonal correspond to linearly independent columns of A.

import numpy as np
linalg = np.linalg

def independent_columns(A, tol = 1e-05):
    """
    Return an array composed of independent columns of A.

    Note the answer may not be unique; this function returns one of many
    possible answers.

    http://stackoverflow.com/q/13312498/190597 (user1812712)
    http://math.stackexchange.com/a/199132/1140 (Gerry Myerson)
    http://mail.scipy.org/pipermail/numpy-discussion/2008-November/038705.html
        (Anne Archibald)

    >>> A = np.array([(2,4,1,3),(-1,-2,1,0),(0,0,2,2),(3,6,2,5)])
    >>> independent_columns(A)
    np.array([[1, 4],
              [2, 5],
              [3, 6]])
    """
    Q, R = linalg.qr(A)
    independent = np.where(np.abs(R.diagonal()) > tol)[0]
    return A[:, independent]

def matrixrank(A,tol=1e-8):
    """
    http://mail.scipy.org/pipermail/numpy-discussion/2008-February/031218.html
    """
    s = linalg.svd(A,compute_uv=0)
    return sum( np.where( s>tol, 1, 0 ) )


matrices = [
    np.array([(2,4,1,3),(-1,-2,1,0),(0,0,2,2),(3,6,2,5)]),
    np.array([(1,2,3),(2,4,6),(4,5,6)]).T,
    np.array([(1,2,3,1),(2,4,6,2),(4,5,6,3)]).T,
    np.array([(1,2,3,1),(2,4,6,3),(4,5,6,3)]).T,
    np.array([(1,2,3),(2,4,6),(4,5,6),(7,8,9)]).T
    ]

for A in matrices:
    B = independent_columns(A)
    assert matrixrank(A) == matrixrank(B) == B.shape[-1]

assert matrixrank(A) == matrixrank(B) 检查independent_columns 函数是否返回与A 具有相同秩的矩阵.

assert matrixrank(A) == matrixrank(B) checks that the independent_columns function returns a matrix of the same rank as A.

assert matrixrank(B) == B.shape[-1] 检查 B 的秩是否等于 B 的列数>.

assert matrixrank(B) == B.shape[-1] checks that the rank of B equals the number of columns of B.

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Covariance