user3789873 - 10 months ago 45

R Question

I have say 25 customers. Each customer has a number of users of our system, e.g. customer 1 has 45 users, customer 2 has 46 users... customer 25 has 1000 users.

I want to bin each customer into a bucket, where each bucket contains a roughly equal number of users. I know that I want 5 buckets in total.

(The buckets here represent servers, I want to apportion my clients to different servers where the total number of users per server is roughly equal, so as to prevent overloading the servers. 1 client has to be on the same server (i.e. can't split 1 client over 2 servers).

Any idea of suitable methods for apportioning customers to buckets? I thought some clustering methods might work (I tried kmeans using R), but I cant seem to find ways of stipulating that the total number of users in each cluster is roughly the same.

Here's my R code as an example of what I've done so far:

`#Create dataset`

r <- data.frame(users=c(1000, 960, 920, 870, 850, 700, 600, 550, 520, 500, 420, 400, 390, 300, 210, 200, 160, 80, 70, 50, 49, 48, 47, 46, 45))

#Try kmeans clustering

fit <- kmeans(r, 5)

#get cluster means

aggregate(r, by=list(fit$cluster),FUN = mean)

#append cluster assignment

r <- data.frame(r,fit$cluster)

#Plot cluster

library(cluster)

clusplot(r, fit$cluster, color=TRUE, shade=TRUE, labels=2, lines=0)

library(fpc)

plotcluster(r, fit$cluster)

This clusters my customers into buckets, but the number of users in each bucket is not roughly equal.

I've tagged this as an R problem, but if there's a simple solution in some other package I'm all ears :-)

Answer Source

I don't know what the recommended solution for such 'constant sum sampling ' is. Here's my shot at it -- sort the items, convert to matrix where each column represents a sample, reverse every other row.

Here's the code:

```
set.seed(1024)
r <- data.frame(users=c(1000, 960, 920, 870, 850, 700, 600, 550, 520, 500, 420, 400, 390, 300, 210, 200, 160, 80, 70, 50, 49, 48, 47, 46, 45))
a<- r$users #runif(n = 25, 100,400) #rnorm(25,100,100) # 1:25
#hist(a)
df<- data.frame(id=1:25,x=a)
# sort
x<- df$id[order(df$x)]
# convert to matrix
#each column of this matrix represetns one sample
xm<-matrix(x,ncol=5,byrow = T); xm
oldsum<-apply(matrix(df$x,ncol=5,byrow = T), 2,sum)
#flip alternate rows of this sorted matrix
i= 1:nrow(xm)
im=i[c(F,T)]
xm[im,]
xm[im,]<- rev(xm[im,])
# new matrix of indeices
xm
#hence the new matrix of values
xm2<- matrix(a[c(xm)],ncol = 5, byrow = F)
xm
xm2
newsum<- (apply(xm2, 2,sum))
# improvement
rbind(oldsum,newsum)
barplot(rbind(oldsum,newsum)[1,])
barplot(rbind(oldsum,newsum)[2,])
# each column of following matrix represents one sample
#(values are indices in original vector a)
xm
```