2
$\begingroup$

Normally when I do factor analysis, I have a whole bunch of variables that need to be reduced. But here I only have two binary variables (yes/no) that I need to reduce into one interval factor. Is Principle Components / Factor Analysis appropriate for this? When I do it, my extraction communalities are really high. I might need a reference to back this up with reviewers.

$\endgroup$
3
$\begingroup$

It is normally considered that three is the minimum number of variables to conduct factor analysis; amongst elsewhere this is maintained in the Wikipedia article (which has a reference) and in some (most? all?) statistical software.

There is no reason however that you can't do principal components analysis (which is not the same as factor analysis, although closely related) to identify which principal component explains most of the variance, even if you only have two binary variables. The correlations between the two can still be calculated.

See for example the below, where Bin1 and Bin3 are correlated binary variables. The first principal component explains most of the variance, and naturally is equally weighted on both of the original variables.

> eg <- data.frame(
+ Bin1 =sample(c(0,1),1000, replace=TRUE),
+ Bin2 =sample(c(0,1),1000, replace=TRUE))
> 
> eg$Bin3 <- ifelse(runif(1000)>.2, eg$Bin1, eg$Bin2)
> cor(eg)
            Bin1        Bin2      Bin3
Bin1  1.00000000 -0.05206971 0.8081088
Bin2 -0.05206971  1.00000000 0.1404252
Bin3  0.80810881  0.14042523 1.0000000


> mod <- princomp(eg[,c("Bin1", "Bin3")])
> ld <- mod$loadings
> attach(eg)
> plot(jitter(Bin1), jitter(Bin3), bty="l", 
main="Jittered version of binary data,\nwith first principal component shown")
> grid()
> lines(Bin1, ld[2,1]/ld[1,1] * (Bin1-mean(Bin1)) + mean(Bin3), col="red")

The scatterplot of the two binary correlated variables (points are jittered):

enter image description here

> summary(mod)
Importance of components:
                          Comp.1     Comp.2
Standard deviation     0.6722961 0.21901598
Proportion of Variance 0.9040544 0.09594559
Cumulative Proportion  0.9040544 1.00000000
> par(mfrow=c(1,2))
> plot(mod)
> biplot(mod)

Component eigenvalues, left, and biplot (loadings+scores), right:

enter image description here

$\endgroup$
4
  • $\begingroup$ I provided titles for your plots, please check. The line on the scatterplot seems to be a regression line (right?); if so it might be mistakenly taken as the 1st PC. $\endgroup$
    – ttnphns
    Feb 22 '13 at 22:26
  • $\begingroup$ @Grrr, in addition to Peter's answer you might want to read my opinion here (with further link there) $\endgroup$
    – ttnphns
    Feb 22 '13 at 22:33
  • $\begingroup$ Yes, it's the regression line. If I get round to it I will change it to the first principal component. Of course in this case it will be very similar, due to the symmetrical nature of the data. $\endgroup$ Feb 23 '13 at 1:53
  • $\begingroup$ Now it's been changed from a regression line to the 1st PC. I nicked some code from a very useful blog on this topic at cerebralmastication.com/2010/09/… $\endgroup$ Feb 23 '13 at 4:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.