Reputation: 81
I created two clusters using the k-means algorithm. Each cluster contains 4 variables. If I want to get the means of each variables in each cluster, should I do:
clusteredsubset$centers
or
colMeans(y[clusteredsubset$cluster == 1,])
colMeans(y[clusteredsubset$cluster == 2,])
where y
is the data matrix (4 columns) and clusteredsubset
is the result of kmeans
.
Upvotes: 1
Views: 3184
Reputation: 73385
Either one is fine, as they give the same result. But since kmeans
returns centers
, why not use it?
The following is based on the first example from ?kmeans
:
set.seed(0)
x <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2),
matrix(rnorm(100, mean = 1, sd = 0.3), ncol = 2))
colnames(x) <- c("x", "y")
cl <- kmeans(x, 2)
## what `kmeans` returns
cl$centers
# x y
#1 -0.0008158201 -0.008394296
#2 0.9261878482 1.029984748
## manual computation
colMeans(x[cl$cluster == 1, ])
# x y
#-0.0008158201 -0.0083942957
colMeans(x[cl$cluster == 2, ])
# x y
#0.9261878 1.0299847
The results are exactly the same (the difference in number of digits is just a printing effect).
## make a plot
plot(x, col = cl$cluster)
points(cl$centers, col = 1:2, pch = 8, cex = 2)
Upvotes: 3
Reputation: 114
I would use:
means = sapply(split(clusteredsubset, clusteredsubset$cluster),function(x)
{return(sapply(x,function(x){return(mean(x))}))})
Upvotes: 1