R help - Apr 2003 - plot clusters

Hi,

I have a dataset which has more than two clusters (say 3 clusters).
I used kmeans to cluster the dataset.

I am wondering how I can plot the clustering result on a two-dimensional
figure????

The example in the kmeans help file is as follows:

 x <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2),
                matrix(rnorm(100, mean = 1, sd = 0.3), ncol = 2))
     cl <- kmeans(x, 2, 20)
     plot(x, col = cl$cluster)

It seems that this is only working for 2 clusters. When my dataset has more
than two clusters, the plot results are obvious wrong.

Any suggestions to plot more than two clusters.

Thanks!!!

>>>>> "pingzhao" == pingzhao  <pingzhao at
waffle.cs.dal.ca>
>>>>>     on Fri, 25 Apr 2003 16:45:19 -0300 writes:
    pingzhao> Hi,
    pingzhao> I have a dataset which has more than two clusters (say 3
clusters).
    pingzhao> I used kmeans to cluster the dataset.

    pingzhao> I am wondering how I can plot the clustering result on a
two-dimensional
    pingzhao> figure????

    pingzhao> The example in the kmeans help file is as follows:

    pingzhao> x <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2),
    pingzhao>            matrix(rnorm(100, mean = 1, sd = 0.3), ncol = 2))
    pingzhao> cl <- kmeans(x, 2, 20)
    pingzhao> plot(x, col = cl$cluster)

    pingzhao> It seems that this is only working for 2 clusters. When my
dataset has more
    pingzhao> than two clusters, the plot results are obvious wrong.

Huh?!
Probably the clustering (method) wasn't what you expected, 
but the plot results are obviously correct!

 x3 <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2),
	     matrix(rnorm(100, mean = 1,  sd = 0.3), ncol = 2),
	     matrix(rnorm(100, mean = 1.8, sd = 0.3), ncol = 2))
 cl3 <- kmeans(x3, 3, 20)
 plot(x3, col = cl3$cluster)

## or even more clear, and also usable on black & white :

 plot(x3, col = cl3$cluster, pch = cl3$cluster)

-----

The much more interesting question is what to do when `x' has
more than two *dimensions*.
The easiest is to then plot on the first two principal
components, but I'm sure Christian Hennig will tell about much
more sophisticated solutions coming Monday...

Regards,
Martin