问题描述
在R中使用princomp()函数时，遇到以下错误:"covariance matrix is not non-negative definite".
While using princomp() function in R, the following error is encountered : "covariance matrix is not non-negative definite".
我认为，这是由于协方差矩阵中的某些值为零(实际上接近零，但在舍入过程中变为零)所致.
I think, this is due to some values being zero (actually close to zero, but becomes zero during rounding) in the covariance matrix.
当协方差矩阵包含零时，是否可以进行PCA处理?
Is there a work around to proceed with PCA when covariance matrix contains zeros ?
[FYI:获得协方差矩阵是princomp()调用内的中间步骤.可以从此处下载用于重现此错误的数据文件-http://tinyurl.com/6rtxrc3]
[FYI : obtaining the covariance matrix is an intermediate step within the princomp() call. Data file to reproduce this error can be downloaded from here - http://tinyurl.com/6rtxrc3]
推荐答案
第一个策略可能是减少公差参数.在我看来，princomp不会传递公差参数，但是prcomp确实接受"tol"参数.如果无效，则应确定协方差几乎为零的向量:
The first strategy might be to decrease the tolerance argument. Looks to me that princomp won't pass on a tolerance argument but that prcomp does accept a 'tol' argument. If not effective, this should identify vectors which have nearly-zero covariance:
nr0=0.001
which(abs(cov(M)) < nr0, arr.ind=TRUE)
这将识别具有负特征值的向量:
And this would identify vectors with negative eigenvalues:
which(eigen(M)$values < 0)
使用help(qr)页面上的h9示例:
Using the h9 example on the help(qr) page:
> which(abs(cov(h9)) < .001, arr.ind=TRUE)
row col
[1,] 9 4
[2,] 8 5
[3,] 9 5
[4,] 7 6
[5,] 8 6
[6,] 9 6
[7,] 6 7
[8,] 7 7
[9,] 8 7
[10,] 9 7
[11,] 5 8
[12,] 6 8
[13,] 7 8
[14,] 8 8
[15,] 9 8
[16,] 4 9
[17,] 5 9
[18,] 6 9
[19,] 7 9
[20,] 8 9
[21,] 9 9
> qr(h9[-9,-9])$rank
[1] 7 # rank deficient, at least at the default tolerance
> qr(h9[-(8:9),-(8:9)])$ take out only the vector with the most dependencies
[1] 6 #Still rank deficient
> qr(h9[-(7:9),-(7:9)])$rank
[1] 6
另一种方法可能是使用alias函数:
Another approach might be to use the alias function:
alias( lm( rnorm(NROW(dfrm)) ~ dfrm) )
这篇关于当协方差矩阵为零时，如何在R中使用princomp()函数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！