Univariate Model Based Clustering

Steve Simon

2006-04-18

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[StATS]: Univariate Model Based Clustering (April 18, 2006).

Back in 2001, I attended an excellent short course on a new approach to cluster analysis taught by Adrian Raftery and Chris Fraley at the Joint Statistics Meetings. Their approach, model based clustering, examined the fits of mixtures of normal distributions. This approach is useful for unidimensional and multidimensional data and has many advantages over other clustering approaches like hierarchical clustering and k-means clustering.

While I greatly enjoyed that class, I never had need to use this approach until just recently. So I dusted off my old notes and worked out a few simple examples to refresh my memory. I want to present some of these examples here on this weblog.

The univariate mixture of normal distributions is very simple and easy to display the results. I downloaded a file from the DASL

associated with the Clustering Cars Story. this file includes measurements on 38 1978-79 model automobiles. The data file has the following variables:

I wanted to look first at the mpg column by itself. Here is a rug plot of the data.

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Notice that there are some gaps in the data that might indicate that the mpg data comes from a mixture of two or possibly three different distributions. In fact, the DASL website alludes to three possible groups of cars:

large sedans (Ford LTD, Chevrolet Caprice Classic), compact cars (Datsun 210, Chevrolet Chevette), and upscale, but smaller, sedans (BMW 320i, Audi 5000).

You can run model based clustering using the mclust library of R. Here’s the code to get the data into R and to draw the rug plot.

library(mclust) cars.data <- read.csv("cars.csv",header=T,as.is=T) par(mar=c(2.1,0.1,0.1,0.1)) plot(range(cars.dataMPG),c(0,0.15),type="n",xlab=" ",ylab=" ",axes=F) axis(side=1) segments(cars.data$MPG,0,cars.data$MPG,0.15)

When you fit a mixture of univariate normal distributions, you have the option of mixing normal distributions with equal variances (the E option in mclust) or allowing each distribution in the normal mixture to have its own variance (the V option in mclust). Here is the sequence of mixtures from 1 to 6 assuming equal variances (E option).

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The code that created these plots is shown below.

par(mar=c(2.1,0.1,0.1,0.1)) mpg.mclust.e <- EMclust(cars.dataMPG,emModelNames="E") x0 <- seq(min(cars.data$MPG),max(cars.data$MPG),length.out=1000) e <- summary(mpg.mclust.e,data=cars.data$MPG,G=1) nt <- rep(0,1000) plot(range(cars.data$MPG),c(0,0.15),type="n",xlab=" ",ylab=" ",axes=F) axis(side=1) nt <- dnorm(x0,mean=e$mu,sd=sqrt(e$sigmasq)) lines(x0,nt,lwd=3) segments(cars.data$MPG,0,cars.data$MPG,0.02) for (k in 2:6) { e <- summary(mpg.mclust.e,data=cars.data$MPG,G=k) nt <- rep(0,1000) plot(range(cars.data$MPG),c(0,0.15),type="n",xlab=" ",ylab=" ",axes=F) axis(side=1) for (i in 1:k) { ni <- e$pro[i]*dnorm(x0,mean=e$mu[i],sd=sqrt(e$sigmasq)) lines(x0,ni,col=9) nt <- nt+ni } lines(x0,nt,lwd=3) segments(cars.data$MPG,0,cars.data$MPG,0.02) }

Compare this to the sequence of fits when you allow the variances to be unique in each component of the normal mixture (V option).

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Here’s the code that produced these graphs:

mpg.mclust.v <- EMclust(cars.dataMPG,emModelNames="V") x0 <- seq(min(cars.data$MPG),max(cars.data$MPG),length.out=1000) v <- summary(mpg.mclust.v,data=cars.data$MPG,G=1) nt <- rep(0,1000) plot(range(cars.data$MPG),c(0,0.15),type="n",xlab=" ",ylab=" ",axes=F) axis(side=1) ni <- dnorm(x0,mean=v$mu,sd=sqrt(v$sigmasq)) lines(x0,ni,lwd=3) segments(cars.data$MPG,0,cars.data$MPG,0.02) for (k in 2:6) { v <- summary(mpg.mclust.v,data=cars.data$MPG,G=k) nt <- rep(0,1000) plot(range(cars.data$MPG),c(0,0.15),type="n",xlab=" ",ylab=" ",axes=F) axis(side=1) for (i in 1:k) { ni <- v$pro[i]*dnorm(x0,mean=v$mu[i],sd=sqrt(v$sigmasq[i])) lines(x0,ni,col=9) nt <- nt+ni } lines(x0,nt,lwd=3) segments(cars.data$MPG,0,cars.data$MPG,0.02) }

Notice that adding another component to the normal mixture does not always result in adding a new mode to the data. For example, under the three component model with the equal variances option, the largest two components are so close together that there is not a distinct separation between the two distributions. Instead, the larger normal distribution turns the nice bell shaped curve into a hunchback.

also notice that with the V option, you see a mixture of both narrow and broad bell curves. Notice, in particular, the huge shifts in variation in the three component model.

Whether you choose a common variance for each component or allow a different variance is a choice that depends a lot on the context of the problem. In this particular application, we have no reason to believe that economy cars should have the same variation as large sedans, so the V option looks more attractive. The disadvantage is that you have to estimate a larger number of parameters in this model.

How many components should there be in the normal mixture? That’s a hard call. The large gap around 25 seems to indicate that there are at least two different groups here, and a similar but smaller gap might be present around 33 mpg or possibly even 19 mpg. So we should give some consideration to there being three or even four clusters.

The mclust library offers a statistic, BIC, that can help us choose a rational number of mixture components. BIC stands for Bayes Information Criterion, and it examines the likelihood under the various models and it has a pently for models that are unduly complex. You can also use the BIC to decide whether to use the simpler equal variances mixture (E option) or the more flexible but more complex V option.

Here is a plot of the BIC values

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and here is a table of those values

E V 1 -256.9093 -256.9093 2 -247.1326 -247.9348 3 -249.9914 -253.2358 4 -256.2769 -256.2545 5 -256.6972 -265.9905 6 -263.8528 -275.1297 7 -264.7732 NA 8 -268.0281 NA 9 -274.4309 NA

I find that these BIC values are hard to read, so I look at how far each value is from the maximum and round to the nearest integer.

E V 1 10 10 2 0 1 3 3 6 4 9 9 5 10 19 6 17 28 7 18 NA 8 21 NA 9 27 NA

Notice that the program refused to fit 7 or more components with the V option, because there is not enough data to estimate the mean and standard deviation of all those components. As a general rule, you want to choose the largest BIC (in this case, the BIC that is the least negative), but anything within 2 units of the best BIC is still a serious competitor. Any BIC value more than 10 units away is not a credible alternative. Using those criteria, we have fairly strong evidence of two components in this normal mixture, and there is little difference between the E and V options in a two component model. A three component model might merit some consideration, but a single component or four component model is about at our limits of credibility.

Notice that the BIC values for the E and V options are identical at the single component model. This just reminds you that allowing different variances for each component is essentially the same as forcing the variances to be equal when there is just a single component.

Creative CommonsLicense This work is licensed under a Creative Commons Attribution 3.0 United States License. It was written by Steve Simon and was last modified on 04/01/2010.

This page was written by Steve Simon while working at Children’s Mercy Hospital. Although I do not hold the copyright for this material, I am reproducing it here as a service, as it is no longer available on the Children’s Mercy Hospital website. Need more information? I have a page with general help resources. You can also browse for pages similar to this one at

for pages similar to this one at with general help resources. You can also browse Children’s Mercy Hospital website. Need more information? I have a page reproducing it here as a service, as it is no longer available on the Hospital. Although I do not hold the copyright for this material, I am This page was written by Steve Simon while working at Children’s Mercy