Introduction

Here I describe two different solutions to decide which number of components to sue for PCA with non-graphical solutions to teh scree test:

Kaiser Rule (aka Optimal Coordinate) \(n_{oc}\). In its simplest form, the Kaiser’s rule retains only the PCs with variances exceeding 1. If a PC has less variance than 1, it means that it explains less total variance than a single variable in the data, which makes it useless.
Acceleration Factor. For every \(j\)-th eigenvalue, the acceleration factor \(a\) is calculated as the change in the slope between the line connecting the \(eig_j\) and \(eig_{j-1}\), and the line connecting \(eig_j\) and \(eig_{j+1}\) \[ a_{j} = (eig_{j+1} - eig_{j}) - (eig_{j} - eig_{j-1}) \] Once the largest \(a_j\) is found, the number of components is set to \(j-1\).

Learn by coding

# Prepare environment ----------------------------------------------------------

library(nFactors)
library(psych)

# Perform PCA
res <- psych::pca(Harman.5)

# Extract eigenvalues
eigenvalues <- res$values

# Graph
plotuScree(x = eigenvalues)

# Non-graphical solutions
ngs <- nScree(x = eigenvalues)

# Kaiser rule\
nkaiser_man <- sum(eigenvalues > 1)

# Accelration factor
a <- NULL
for (j in 2:(length(eigenvalues) - 1)){
  a[j] <- (eigenvalues[j + 1] - eigenvalues[j]) - (eigenvalues[j] - eigenvalues[j - 1])
}

naf_man <- which.max(a) - 1

# Compare results
data.frame(manual = c(naf = naf_man, nkaiser = nkaiser_man),
           nFactor = c(naf = ngs$Components[["naf"]],
                       nkaiser = ngs$Components[["nkaiser"]]))

        manual nFactor
naf          2       2
nkaiser      2       2

TL;DR, just give me the code!

# Prepare environment ----------------------------------------------------------

library(nFactors)
library(psych)

# Perform PCA
res <- psych::pca(Harman.5)

# Extract eigenvalues
eigenvalues <- res$values

# Graph
plotuScree(x = eigenvalues)

# Non-graphical solutions
ngs <- nScree(x = eigenvalues)

# Kaiser rule\
nkaiser_man <- sum(eigenvalues > 1)

# Accelration factor
a <- NULL
for (j in 2:(length(eigenvalues) - 1)){
  a[j] <- (eigenvalues[j + 1] - eigenvalues[j]) - (eigenvalues[j] - eigenvalues[j - 1])
}

naf_man <- which.max(a) - 1

# Compare results
data.frame(manual = c(naf = naf_man, nkaiser = nkaiser_man),
           nFactor = c(naf = ngs$Components[["naf"]],
                       nkaiser = ngs$Components[["nkaiser"]]))

Deciding the Number of PCs with Non-Graphical Solutions to the Scree Test

Introduction

Learn by coding

TL;DR, just give me the code!

Other resources