Reputation: 49
I tried to use princomp()
and principal()
to do PCA in R with data set USArressts. However, I got two different results for loadings/rotaion and scores.
First, I centered and normalised the original data frame so it is easier to compare the outputs.
library(psych)
trans_func <- function(x){
x <- (x-mean(x))/sd(x)
return(x)
}
A <- USArrests
USArrests <- apply(USArrests, 2, trans_func)
princompPCA <- princomp(USArrests, cor = TRUE)
principalPCA <- principal(USArrests, nfactors=4 , scores=TRUE, rotate = "none",scale=TRUE)
Then I got the results for the loadings and scores using the following commands:
princompPCA$loadings
principalPCA$loadings
Could you please help me to explain why there is a difference? and how can we interprete these results?
Upvotes: 1
Views: 2728
Reputation: 474
ev <- eigen(R) # R is a correlation matrix of DATA
ev$vectors %*% diag(ev$values) %*% t(ev$vectors)
pc <- princomp(scale(DATA, center = F, scale = T),cor=TRUE)
p <-principal(DATA, rotate="none")
#eigen values
ev$values^0.5
pc$sdev
p$values^0.5
#eigen vectors - loadings
ev$vectors
pc$loadings
p$weights %*% diag(p$values^0.5)
pc$loading %*% diag(pc$sdev)
p$loadings
#weights
ee <- diag(0,2)
for (j in 1:2) {
for (i in 1:2) {
ee[i,j] <- ev$vectors[i,j]/p$values[j]^0.5
}
};ee
#scores
s <- as.matrix(scale(DATA, center = T, scale = T)) %*% ev$vectors
scale(s)
p$scores
scale(pc$scores)
Upvotes: 0
Reputation: 79
In 4 years, I would like to provide a more accurate answer to this question. I use iris data as an example.
data = iris[, 1:4]
First, do PCA by the eigen-decomposition
eigen_res = eigen(cov(data))
l = eigen_res$values
q = eigen_res$vectors
Then the eigenvector corresponding to the largest eigenvalue is the factor loadings
q[,1]
We can treat this as a reference or the correct answer. Now we check the results by different r functions. First, by function 'princomp'
res1 = princomp(data)
res1$loadings[,1]
# compare with
q[,1]
No problem, this function actually just return the same results as 'eigen'. Now move to 'principal'
library(psych)
res2 = principal(data, nfactors=4, rotate="none")
# the loadings of the first PC is
res2$loadings[,1]
# compare it with the results by eigendecomposition
sqrt(l[1])*q[,1] # re-scale the eigen vector by sqrt of eigen value
You may find they are still different. The problem is the 'principal' function does eigendecomposition on the correlation matrix by default. Note: PCA is not invariant with rescaling the variables. If you modify the code as
res2 = principal(data, nfactors=4, rotate="none", cor="cov")
# the loadings of the first PC is
res2$loadings[,1]
# compare it with the results by eigendecomposition
sqrt(l[1])*q[,1] # re-scale the eigen vector by sqrt of eigen value
Now, you will get the same results as 'eigen' and 'princomp'.
Summarize:
Upvotes: 1
Reputation: 79
At the very end of the help document of ?principal
:
"The eigen vectors are rescaled by the
sqrt
of the eigen values to produce the component loadings more typical in factor analysis."
So principal
returns the scaled loadings. In fact, principal
produces a factor model estimated by the principal component method.
Upvotes: 3