问题描述
我正在使用numpy.cov从超过400个时间序列的数据集中创建协方差矩阵.使用linalg.det给我零值,所以矩阵是奇异的.我可以使用linalg.svd看到等级比列数少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.我在基础时间序列上使用了corrcoef,但是没有> 0.78的相关性,因此在那儿不明显.有人可以建议一种方法来确定简并列的位置.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,则沿对角线的R
列的值非零. A
的独立列.
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
相同等级的矩阵.
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
的等级是否等于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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