I have performed PCA using prcomp in R with my databases of 75-76 indicator variables and 7232 companies, including NAs. Before applying the function, I centred my data, but did not rescale them because they are all indicator variables. (Is my reasoning correct?)
After that I varimax-rotated the loadings of the 2 or 3 first principal components following the instructions by amoeba here.
Since I had centred, but not rescaled my data, I changed the code to:
Varimax_results <- varimax(rawLoadings,normalize = FALSE)
invLoadings <- t(pracma::pinv(VarimaxLoadings))
scores <- scale(DatosPCA, scale = FALSE) %*% invLoadings
Now I am trying to figure out why the scores given by "prcomp" and the scores obtained using the code above are not the same.
I am probably missing some theoretical background, so I would be grateful if someone could tell me if the scores are supposed to be the same and, in that case, what I am doing wrong in my code. If they are not supposed to be the same, which ones should I use?
Thank you very much!
Related
I'm trying to understand the definition of scale that R provides. I have data (mydata) that I want to make a heat map with, and there is a VERY strong positive skew. I've created a heatmap with a dendrogram for both scale(mydata) and log(my data), and the dendrograms are different for both. Why? What does it mean to scale my data, versus log transform my data? And which would be more appropriate if I want to look at the dendrogram illustrating the relationship between the columns of my data?
Thank you for any help! I've read the definitions but they are whooping over my head.
log simply takes the logarithm (base e, by default) of each element of the vector.
scale, with default settings, will calculate the mean and standard deviation of the entire vector, then "scale" each element by those values by subtracting the mean and dividing by the sd. (If you use scale(x, scale=FALSE), it will only subtract the mean but not divide by the std deviation.)
Note that this will give you the same values
set.seed(1)
x <- runif(7)
# Manually scaling
(x - mean(x)) / sd(x)
scale(x)
It provides nothing else but a standardization of the data. The values it creates are known under several different names, one of them being z-scores ("Z" because the normal distribution is also known as the "Z distribution").
More can be found here:
http://en.wikipedia.org/wiki/Standard_score
This is a late addition but I was looking for information on the scale function myself and though it might help somebody else as well.
To modify the response from Ricardo Saporta a little bit.
Scaling is not done using standard deviation, at least not in version 3.6.1 of R, I base this on "Becker, R. (2018). The new S language. CRC Press." and my own experimentation.
X.man.scaled <- X/sqrt(sum(X^2)/(length(X)-1))
X.aut.scaled <- scale(X, center = F)
The result of these rows are exactly the same, I show it without centering because of simplicity.
I would respond in a comment but did not have enough reputation.
I thought I would contribute by providing a concrete example of the practical use of the scale function. Say you have 3 test scores (Math, Science, and English) that you want to compare. Maybe you may even want to generate a composite score based on each of the 3 tests for each observation. Your data could look as as thus:
student_id <- seq(1,10)
math <- c(502,600,412,358,495,512,410,625,573,522)
science <- c(95,99,80,82,75,85,80,95,89,86)
english <- c(25,22,18,15,20,28,15,30,27,18)
df <- data.frame(student_id,math,science,english)
Obviously it would not make sense to compare the means of these 3 scores as the scale of the scores are vastly different. By scaling them however, you have more comparable scoring units:
z <- scale(df[,2:4],center=TRUE,scale=TRUE)
You could then use these scaled results to create a composite score. For instance, average the values and assign a grade based on the percentiles of this average. Hope this helped!
Note: I borrowed this example from the book "R In Action". It's a great book! Would definitely recommend.
I calculated polychoric correlation matrices for the same data frame (20 ordinal variables, 190 missing values) in R, using three different packages and the coefficients for same variables are slightly different from each other.
I used the lavCor function from "lavaan" (I did list the ordinal variables when calling the function), polychoric function from "psych" (1.9.1) (took the rhos), and cor_auto function from "qgraph" (which is supposed to automatically calculate polychoric correlations for ordinal data). I am confused because I thought they were supposed to give exactly the same results. I read package documentations but could not find anything that helped me understand why. Could anyone let me know why this happens? I am sure I am missing some tiny difference between those, but I cannot figure it out.
PS: I guess this could have happened because psych package adjusts missing values (I have 190) using the correction for continuity, but I still do not understand why qgraph yields different results than lavaan as qgraph says it uses lavaan's lavCor function to calculate polychoric correlations.
Thanks!!
depanx<-data[1:20]
cor.depanx<-cor_auto(depanx)
polychor<-polychoric(depanx)
polymat<-polychor$rho
lav<-lavCor(depanx,ordered=c("unh","enj","trd","rst","noG","cry","cnc","htd","bdp","lnl","lov",
"cmp","wrg","pst","sch","dss","hlt","bad","ftr","oth"))
# as a result, matrices "cor.depanx", "polymat", and "lav" are different from each other.
Nice question! I do not know what the "data" dataset in you example is, but i recreate the two possible scenarios, which have most probably caused the discrepancy between cor_auto and lavCor results. In summary, first you must set the "ordinalLevelMax" argument in cor_auto based on your data and second you need to synchronize the "missing" argument in the two functions. Detailed explanation in the code snippet below:
depanx<-data.frame(lapply(1:5,function(x)sample(1:6,100,replace = T)),
stringsAsFactors = F)
colnames(depanx)=LETTERS[1:5]
lav<-lavaan::lavCor(depanx,ordered=colnames(depanx))
cor.depanx<-cor_auto(depanx)
all(lav==cor.depanx)#TRUE
#The first argument in cor auto, which you need to pay attention to is
#"ordinalLevelMax". #It is set to 7 by default in cor_auto,
#so any variable with levels more than 7 is sent to lavCor as plain numeric and not
#ordinal.
#Now we create the same dataset with 8 level variables. lavCor detects all as ordinal,
#since we have labeled them as so by "ordered" argument of lavCor, so it uses
#ploychorial
#correlations. Since "ordinalLevelMax" in cor_auto is 7 by default and you have not
#changed it,
#cor_auto detect none as ordinaland does not send them to lavCor as Ordinalvariables,
#so Lavcor computes pearson correlations between them,all.
depanx2<-data.frame(lapply(1:5,function(x)sample(1:8,100,replace =T)),
stringsAsFactors = F)
colnames(depanx2)=LETTERS[1:5]
lav2<-lavaan::lavCor(depanx2,ordered=colnames(depanx2))
cor.depanx2<-cor_auto(depanx2)
all(lav2==cor.depanx2)#FALSE
# the next argument you must synchronise in lavCor and cor_auto is the "missing",
#which is by default set to "pairwise" and "listwise" in cor_auto and lavCor,
#respectively.
#here we set row 10:20 value of the fifth variable to NA, without synchronizing the
#argument
depanx3<-data.frame(lapply(1:5,function(x)sample(1:6,100,replace =T)),
stringsAsFactors = F)
colnames(depanx3)=LETTERS[1:5]
depanx3[10:20,5]<-NA
lav3<-lavaan::lavCor(depanx3,ordered=colnames(depanx3))
cor.depanx3<-cor_auto(depanx3)
all(lav3==cor.depanx3)#FALSE
I'm relatively new to R and am currently in the process of constructing a PLS model using the pls package. I have two independent datasets of equal size, the first is used here for calibrating the model. The dataset comprises of multiple response variables (y) and 101 explanatory variables (x), for 28 observations. The response variables, however, will each be included seperately in a PLS model. The code current looks as follows:
# load data
data <- read.table("....txt", header=TRUE)
data <- as.data.frame(data)
# define response variables (y)
HEIGHT <- as.numeric(unlist(data[2]))
FBM <- as.numeric(unlist(data[3]))
N <- as.numeric(unlist(data[4]))
C <- as.numeric(unlist(data[5]))
CHL <- as.numeric(unlist(data[6]))
# generate matrix containing the explanatory (x) variables only
spectra <-(data[8:ncol(data)])
# calibrate PLS model using LOO and 20 components
library(pls)
refl.pls <- plsr(N ~ as.matrix(spectra), ncomp=20, validation = "LOO", jackknife = TRUE)
# visualize RMSEP -vs- number of components
plot(RMSEP(refl.pls), legendpos = "topright")
# calculate explained variance for x & y variables
summary(refl.pls)
I have currently arrived at the point at which I need to decide, for each response variable, the optimal number of components to include in my PLS model. The RMSEP values already provide a decent indication. However, I would also like to base my decision on the PRESS (Predicted Residual Sum of Squares) statistic, in accordance various studies comparable to the one I am conducting. So in short, I would like to extract the PRESS statistic for each PLS model with n components.
I have browsed through the pls package documentation and across the web, but unfortunately have been unable to find an answer. If there is anyone out here that could help me get in the right direction that would be greatly appreciated!
You can find the PRESS values in the mvr object.
refl.pls$validation$PRESS
You can see this either by exploring the object directly with str or by perusing the documentation more thoroughly. You will notice if you look at ?mvr you will see the following:
validation if validation was requested, the results of the
cross-validation. See mvrCv for details.
Validation was indeed requested so we follow this to ?mvrCv where you will find:
PRESS a matrix of PRESS values for models with 1, ...,
ncomp components. Each row corresponds to one response variable.
I have conducted an NMDS analysis and have plotted the output too. However, I am unsure how to actually report the results from R. Which parts from the following output are of most importance? The graph that is produced also shows two clear groups, how are you supposed to describe these results?
MDS.out
Call:
metaMDS(comm = dgge2, distance = "bray")
global Multidimensional Scaling using monoMDS
Data: dgge2
Distance: bray
Dimensions: 2
Stress: 0
Stress type 1, weak ties
No convergent solutions - best solution after 20 tries
Scaling: centring, PC rotation, halfchange scaling
Species: expanded scores based on ‘dgge2’
The most important pieces of information are that stress=0 which means the fit is complete and there is still no convergence. This happens if you have six or fewer observations for two dimensions, or you have degenerate data. You should not use NMDS in these cases. Current versions of vegan will issue a warning with near zero stress. Perhaps you had an outdated version.
I think the best interpretation is just a plot of principal component. yOu can use plot and text provided by vegan package. Here I am creating a ggplot2 version( to get the legend gracefully):
library(vegan)
library(ggplot2)
data(dune)
ord = metaMDS(comm = dune)
ord_spec <- scores(ord, "spec")
ord_spec <- cbind.data.frame(ord_spec,label=rownames(ord_spec))
ord_sites <- scores(ord, "sites")
ord_sites <- cbind.data.frame(ord_sites,label=rownames(ord_sites))
ggplot(data=ord_spec,aes(x=NMDS1,y=NMDS2)) +
geom_text(aes(label=label,col='species')) +
geom_text(data=ord_sites,aes(label=label,col='sites'))
I am trying to generate a random set of numbers that exactly mirror a data set that I have (to test it). The dataset consists of 5 variables that are all correlated with different means and standard deviations as well as ranges (they are likert scales added together to form 1 variable). I have been able to get mvrnorm from the MASS package to create a dataset that replicated the correlation matrix with the observed number of observations (after 500,000+ iterations), and I can easily reassign means and std. dev. through z-score transformation, but I still have specific values within each variable vector that are far above or below the possible range of the scale whose score I wish to replicate.
Any suggestions how to fix the range appropriately?
Thank you for sharing your knowledge!
To generate a sample that does "exactly mirror" the original dataset, you need to make sure that the marginal distributions and the dependence structure of the sample matches those of the original dataset.
A simple way to achieve this is with resampling
my.data <- matrix(runif(1000, -1, 2), nrow = 200, ncol = 5) # Some dummy data
my.ind <- sample(1:nrow(my.data), nrow(my.data), replace = TRUE)
my.sample <- my.data[my.ind, ]
This will ensure that the margins and the dependence structure of the sample (closely) matches those of the original data.
An alternative is to use a parametric model for the margins and/or the dependence structure (copula). But as staded by #dickoa, this will require serious modeling effort.
Note that by using a multivariate normal distribution, you are (implicity) assuming that the dependence structure of the original data is the Gaussian copula. This is a strong assumption, and it would need to be validated beforehand.