Subscript out of bound error in R - r

Question

While Using factanal function from stats package for performing factor analysis.
I tried following thing.
library(mirt)
library(ltm)
library(psych)
library(stats)
data(SAT12)
data=SAT12
cor_mat=polychoric(data, ML=TRUE, global=F)
fit <- factanal(factors=2, n.obs=nrow(data), covmat=cor_mat$rho)
Divide_item_Factor_Loading(fit)
when I am trying to run Divide_item_Factor_Loading(fit) an error called
Error in a[[i]][[2]] : subscript out of bounds
pops up.
my complete code of Divide_item_Factor_Loading:
Divide_item_Factor_Loading=function(fit)
{
a=list()
items=NULL
for(i in 1:nrow(fit$loadings)) ######corresponding to rows of loading matrix
{
k=which(fit$loadings[i,]==max(abs(fit$loadings[i,])))
a[[i]]=c(i,as.numeric(k))
}
fact_item_mat=matrix(, nrow=nrow(fit$loadings), ncol=ncol(fit$loadings))
for(j in 1:(ncol(fit$loadings)))
{
for(i in 1:(nrow(fit$loadings)))
{
if(a[[i]][[2]]==j) {fact_item_mat[i,j]=a[[i]][[1]]}
}
}
nam=names(fit$loadings[,1])
factor=list()
for(i in 1:ncol(fit$loadings))
{
factor[[i]]=sort(fact_item_mat[,i], decreasing = FALSE, na.last = NA)
fac=factor[[i]]
fac=nam[fac]
factor[[i]]=fac
}
names(factor)=paste("factor", 1:ncol(fit$loadings), sep="")
return(factor)
}
What steps should I take now to avoid this error?

Answer 1

To change the way the loadings are printed, use the cutoff argument to print.loadings.
Try something like this:
print(fit$loadings, cutoff=0)
The actual matrix contains all the values.
print(loadings(fit), cutoff=0)
Loadings:
Factor1 Factor2
Item 1 0.014 0.418
Item 2 0.130 0.350
Item 3 0.036 0.553
Item 4 0.166 0.294
Item 5 0.990 0.125
Factor1 Factor2
SS loadings 1.025 0.705
Proportion Var 0.205 0.141
Cumulative Var 0.205 0.346
Now extract the maximum loading on each factor, using apply():
apply(loadings(fit), 2, max)
Factor1 Factor2
0.9895743 0.5531770

Answer 2

Running your code and debugging your function (using debug function) I can see why you're having a "subscript out of bound" error:
the 15th element (among other) of your variable a is of length 1 so R is not happy when you're trying to reach a[[15]][2]...
the reason why this element is only of length one instead of 2 is because the maximum absolute value of factor is reached for a negative value and you're asking which value (not absolute) is equal to this maximal absolute value, so the answer is none...
Hence you need to change the line
which(fit$loadings[i,]==max(abs(fit$loadings[i,]))) to which(abs(fit$loadings[i,])==max(abs(fit$loadings[i,])))
and you'll get:
Divide_item_Factor_Loading(fit)
#$factor1
#[1] "Item.1" "Item.4" "Item.6" "Item.7" "Item.8" "Item.9" "Item.10" "Item.11" "Item.13" "Item.14" "Item.15"
#[12] "Item.17" "Item.19" "Item.20" "Item.24" "Item.26" "Item.27" "Item.28" "Item.29"
#$factor2
#[1] "Item.2" "Item.3" "Item.5" "Item.12" "Item.16" "Item.18" "Item.21" "Item.22" "Item.23" "Item.25" "Item.30"
#[12] "Item.31" "Item.32"
Even if the debugged function will now work, I think you should change it because this is more complicated than it should be.
My proposition for an alternative function:
Divide_item_Factor_Loading_v2<-function(fit){
a<-apply(fit$loadings,1,function(facs) which(abs(facs)==max(abs(facs))))
return(list(factor1=names(a)[a==1],factor2=names(a)[a==2]))
}
This gives for your fit object the exact same result as your (debugged) function:
Divide_item_Factor_Loading_v2(fit)
#$factor1
#[1] "Item.1" "Item.4" "Item.6" "Item.7" "Item.8" "Item.9" "Item.10" "Item.11" "Item.13" "Item.14" "Item.15"
#[12] "Item.17" "Item.19" "Item.20" "Item.24" "Item.26" "Item.27" "Item.28" "Item.29"
#$factor2
#[1] "Item.2" "Item.3" "Item.5" "Item.12" "Item.16" "Item.18" "Item.21" "Item.22" "Item.23" "Item.25" "Item.30"
#[12] "Item.31" "Item.32"

Answer 3

Check ?loadings, what you'll find out that there is a cutoff parameter that defines a value that "loadings smaller than this (in absolute value) are suppressed".