R help - Jul 2008 - PCA : Error in eigen(cv,

Hi all,

I am doing bootstrap on a distance matrix, in which samples have been 
drawn with replacement. After that I do PCA on a resulted matrix, and 
these 2 steps are repeated 1000 times.

pca(x) is a vector where I wanted to store all 1000 PCAs; and x is from 
1 to 1000
SampleD is a new matrix after resampling;

I am getting the following error message, which I don't understand:
....
+pca(x)<-princomp(SampleD[i,j])
+ }
Error in eigen(cv, symmetric = TRUE) : infinite or missing values in 'x'

Should I maybe not use a vector, but matrix instead?
Thanks!

-- 
Tanya.


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Hi, yes a loop would work. Maybe there are other options also. Bootstrapping
a PCA with a loop can be done as follows. You will have to adjust the
dimension of the matrices/array according to the dimensions of your data and
according to the values that you want to retrieve from your PCA.

####Start of example

##Iris data
data=iris

##Inspect data
data

##Perform pca on columns 1:4
x=prcomp(data[,1:4])

##Inspect results of pca
summary(x)
summary(x)$importance

trials=10 ##we want 10 bootstraps

##We want to bootstrap pca
##and, say, know some information
##contained in "summary(x)$importance"
##Thus, we create an array that has the
##dimension of "summary(x)$importance"
##and depth of "trials"
importance=array(0,c(3,4,trials))

##Bootstrap the PCA
for(i in 1:trials){
    ##sample IDs of original data rows
    sam=sample(1:length(data[,1]),replace=T)

    ##sample new data from old data
    ##using the data row IDs from above
    newdata=data[c(sam),]

    ##store the results of the 
    ##importance matrix resulting from PCA
    ##performed on columns 1:4 
    ##of our bootstrapped data
    importance[,,i]=summary(prcomp(newdata[,1:4]))$importance
    }
    
##Now we can assess, for example, 
##the mean and SD
##of the accounted variance of the
##first principal component

##mean of the variance accounted for
##by the first PC over all trials
mean(importance[2,1,])

##sd of the variance accounted for
##by the first PC over all trials
sd(importance[2,1,])

####End of example

Best,
Daniel

-------------------------
cuncta stricte discussurus
-------------------------

-----Urspr?ngliche Nachricht-----
Von: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] Im
Auftrag von Tanya Yatsunenko
Gesendet: Monday, June 30, 2008 10:09 PM
An: r-help at r-project.org
Betreff: [R] PCA : Error in eigen(cv,

Hi all,

I am doing bootstrap on a distance matrix, in which samples have been drawn
with replacement. After that I do PCA on a resulted matrix, and these 2
steps are repeated 1000 times.

pca(x) is a vector where I wanted to store all 1000 PCAs; and x is from
1 to 1000
SampleD is a new matrix after resampling;

I am getting the following error message, which I don't understand:
....
+pca(x)<-princomp(SampleD[i,j])
+ }
Error in eigen(cv, symmetric = TRUE) : infinite or missing values in 'x'

Should I maybe not use a vector, but matrix instead?
Thanks!

--
Tanya.


	[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Ok, the mistake was in the pca(x)<-princomp(SampleD[i,j]), should've used
pca(x)<-princomp(SampleD) instead.
Now, is there anyway to keep track of the matrix index, so in the end of all
PCAs, I can tell which score/loading belongs to which sample?
Thanks everyone!

On Mon, Jun 30, 2008 at 9:08 PM, Tanya Yatsunenko <yata25@gmail.com>
wrote:
>  Hi all,
>
> I am doing bootstrap on a distance matrix, in which samples have been drawn
> with replacement. After that I do PCA on a resulted matrix, and these 2
> steps are repeated 1000 times.
>
> pca(x) is a vector where I wanted to store all 1000 PCAs; and x is from 1
> to 1000
> SampleD is a new matrix after resampling;
>
> I am getting the following error message, which I don't understand:
> ....
> +pca(x)<-princomp(SampleD[i,j])
> + }
> Error in eigen(cv, symmetric = TRUE) : infinite or missing values in
'x'
>
> Should I maybe not use a vector, but matrix instead?
> Thanks!
>
> --
> Tanya.
>
>

-- 
Tanya

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