Constructing components from PLSR loadings in R - r

I want to compute the components for a set of variables using the loadings (weights) from a PLSR using the plsr function.
I thought that the components were computed by summing the values of each variable multiplied by the estimated loading (weight).
However, using the output from plsr and doing that doesn't give me the expected values:
Example:
library("pls")
data(oliveoil)
sens.pcr <- plsr(sensory ~ chemical, ncomp = 4, scale = F, data = oliveoil)
Extract loadings/weights:
df <- cbind(sens.pcr$loadings[,1],sens.pcr$loadings[,2],sens.pcr$loadings[,3],sens.pcr$loadings[,4])
One test observation:
firstrow <- oliveoil$chemical[1,]
Extract the components (scores):
scores <- sens.pcr$scores
Do the linear combination:
sum(firstrow*df[,1])
[1] -12.81924
Which is not equal to the first score scores[1,1] = 0.5100166
What is it that I am missing?
Using the sense.pcr$loadings.weigths didn't make any big difference either.

Related

Is calculated factor scores in robCompositions R package, correct?

Please have a look at the factor scores returned from robCompositions package in this example:
data(expenditures)
x <- expenditures
res.rob <- pfa(x, factors=1, score="regression")
according to pfa help, since the covariance is not specified,
the covariance is estimated from isometric log-ratio transformed data internally, but the data used for factor analysis are back-transformed to the "clr" space.
So the clr transformed data obtain as follows:
# ilr transformation
ilr <- function(x){
x.ilr=matrix(NA,nrow=nrow(x),ncol=ncol(x)-1)
for (i in 1:ncol(x.ilr)){
x.ilr[,i]=sqrt((i)/(i+1))*
log(((apply(as.matrix(x[,1:i]), 1, prod))^(1/i))/(x[,i+1]))
}
return(x.ilr)
}
#construct orthonormal basis:
#(matrix with ncol(x) rows and ncol(x)-1 columns)
V=matrix(0,nrow=ncol(x),ncol=ncol(x)-1)
for (i in 1:ncol(V)){
V[1:i,i] <- 1/i
V[i+1,i] <- (-1)
V[,i] <- V[,i]*sqrt(i/(i+1))
}
z=ilr(x) #ilr transformed data
y=z%*%t(V) #clr transformed data
now the factor scores using regression method might be calculated as follows:
loa<-c(0.970,0.830,0.986,0.876,0.977) #res.rob object
facscores<- y%*%loa
head(facscores)
-0.009485110
0.009680645
0.008426665
-0.015401000
-0.003610644
-0.004584145
but calling res.rob$scores returns us
head(res.rob$scores)
Factor1
-755.2681
705.5309
4196.5652
-778.6955
-628.2141
-663.4534
So please check am I wrong or there is probably a bug in the pfa command?
Yours,
Hamid

How to use lapply with get.confusion_matrix() in R?

I am performing a PLS-DA analysis in R using the mixOmics package. I have one binary Y variable (presence or absence of wetland) and 21 continuous predictor variables (X) with values ranging from 1 to 100.
I have made the model with the data_training dataset and want to predict new outcomes with the data_validation dataset. These datasets have exactly the same structure.
My code looks like:
library(mixOmics)
model.plsda<-plsda(X,Y, ncomp = 10)
myPredictions <- predict(model.plsda, newdata = data_validation[,-1], dist = "max.dist")
I want to predict the outcome based on 10, 9, 8, ... to 2 principal components. By using the get.confusion_matrix function, I want to estimate the error rate for every number of principal components.
prediction <- myPredictions$class$max.dist[,10] #prediction based on 10 components
confusion.mat = get.confusion_matrix(truth = data_validatie[,1], predicted = prediction)
get.BER(confusion.mat)
I can do this seperately for 10 times, but I want do that a little faster. Therefore I was thinking of making a list with the results of prediction for every number of components...
library(BBmisc)
prediction_test <- myPredictions$class$max.dist
predictions_components <- convertColsToList(prediction_test, name.list = T, name.vector = T, factors.as.char = T)
...and then using lapply with the get.confusion_matrix and get.BER function. But then I don't know how to do that. I have searched on the internet, but I can't find a solution that works. How can I do this?
Many thanks for your help!
Without reproducible there is no way to test this but you need to convert the code you want to run each time into a function. Something like this:
confmat <- function(x) {
prediction <- myPredictions$class$max.dist[,x] #prediction based on 10 components
confusion.mat = get.confusion_matrix(truth = data_validatie[,1], predicted = prediction)
get.BER(confusion.mat)
}
Now lapply:
results <- lapply(10:2, confmat)
That will return a list with the get.BER results for each number of PCs so results[[1]] will be the results for 10 PCs. You will not get values for prediction or confusionmat unless they are included in the results returned by get.BER. If you want all of that, you need to replace the last line to the function with return(list(prediction, confusionmat, get.BER(confusion.mat)). This will produce a list of the lists so that results[[1]][[1]] will be the results of prediction for 10 PCs and results[[1]][[2]] and results[[1]][[3]] will be confusionmat and get.BER(confusion.mat) respectively.

Different results when performing PCA in R with princomp() and principal ()

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?
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.
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$loadings[,1]
# compare it with the results by eigendecomposition
sqrt(l[1])*q[,1] # re-scale the eigen vector by sqrt of eigen value
Now, you will get the same results as 'eigen' and 'princomp'.
Summarize:
If you want to do PCA, you'd better apply 'princomp' function.
PCA is a special case of the Factor model or a simplified version of the factor model. It is just equivalent to eigendecomposition.
We can apply PCA to get an approximation of a factor model. It doesn't care about the specific factors, i.e. epsilons in a factor model. So, if you change the number of factors in your model, you will get the same estimations of the loadings. It is different from the maximum likelihood estimation.
If you are estimating a factor model, you'd better use 'principal' function, since it provides more functions, like rotation, calculating the scores by different methods, and so on.
Rescale the loadings of a PCA model doesn't affect the results too much. Since you still project the data onto the same optimal direction, i.e. maximize the variation in the resulting PC.
ev <- eigen(R) # R is a correlation matrix of DATA
ev$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)

How to find an optimal adstock decay factor for an independent variable in panel analysis in R?

I'm working with a panel dataset (24 months of data for 210 DMAs). I'm trying to optimize the adstock decay factor for an independent variable by minimizing the standard error of a fixed effects model.
In this particular case, I want to get a decay factor that minimizes the SE of the adstock-transformed variable "SEM_Br_act_norm" in the model "Mkt_TRx_norm = b0 + b1*Mkt_TRx_norm_prev + b2*SEM+Br_act_norm_adstock".
So far, I've loaded the dataset in panel formal using plm and created a function to generate the adstock values. The function also runs a fixed effects model on the adstock values and returns the SE. I then use optimize() to find the best decay value within the bounds (0,1). While my code is returning an optimal value, I am worried something is wrong because it returns the same optimum (close to 1) on all other variables.
I've attached a sample of my data, as well as key parts of my code. I'd greatly appreciate if someone could take a look and see what is wrong.
Sample Data
# Set panel data structure
alldata <- plm.data (alldata, index = c("DMA", "Month_Num"))
alldata$var <- alldata$SEM_Br_act_norm +0
# Create 1 month time lag for TRx
alldata <- ddply(
alldata, .(DMA), transform,
# This assumes that the data is sorted
Mkt_TRx_norm_prev = c(NA,Mkt_TRx_norm[-length(Mkt_TRx_norm)])
)
# Create adstock function and obtain SE of regression
adstockreg <-function(decay, period, data_vector, pool_vector=0){
data_vector <-alldata$var
pool_vector <- alldata$DMA
data2<-data_vector
l<-length(data_vector)
#if no pool apply zero to vector
if(length(pool_vector)==1)pool_vector<-rep(0,l)
#outer loop: extract data to decay from observation i
for( i in 1:l){
x<-data_vector[i]
#inner loop: apply decay onto following observations after i
for(j in 1:min(period,l)){
#constrain decay to same pool (if data is pooled)
if( pool_vector[i]==pool_vector[min(i+j,l)]){data2[(i+j)]<- data2[(i+j)]+(x*(decay)^j)}
}
}
#reduce length of edited data to equal length of initial data
data2<-data2[1:l]
#regression - excludes NA values
alldata <- plm.data (alldata, index = c("DMA", "Month_Num"))
var_fe <- plm(alldata$Mkt_TRx_norm ~ alldata$Mkt_TRx_norm_prev + data2, data = alldata , model = "within", na.action = na.exclude)
se <- summary(var_fe)$coefficients["data2","Std. Error"]
return(se)
}
# Optimize decay for adstock variable
result <- optimize(adstockreg, interval=c(0,1), period = 6)
print(result)

Z-scores rounded to infinity for small p-values in R

I am working with a genome-wide association study dataset, with p-values ranging from 1E-30 to 1. I have an R data frame "data" which includes a variable "p" for the p-values.
I need to perform genomic correction of the p-values, which I am doing using the following code:
p=data$p
Zsq = qchisq(1-p, 1)
lambda = median(Zsq)/0.456
newZsq = Zsq/lambda
Newp = 1-pchisq(newZsq, 1)
In the command on the second line, where I use the qchisq function to convert p-values to z-scores, z-scores for p-values < 1E-16 are being rounded to infinity. This means the p-values for my most significant data points are rounded to 0 after the genomic correction, and I lose their ranking.
Is there any way around this?
Read help(".Machine"). Then set lower.tail=FALSE and avoid taking differences with 1:
p <- 1e-17
Zsq = qchisq(p, 1, lower.tail=FALSE)
lambda = median(Zsq)/0.456
newZsq = Zsq/lambda
Newp = pchisq(newZsq, 1, lower.tail=FALSE)
#[1] 0.4994993

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