Prasanna <prasannaprakash <at> gmail.com> writes:

>

> Dear R experts

>

> I am looking for a package which gives me latin hyper cube samples

> from the grid of values produced from the command "expand.grid". Any

> pointers to this issue might be very useful. Basically, I am doing the

> following:

>

> > a<-(1:10)

> > b<-(20:30)

> > dataGrid<-expand.grid(a,b)

>

> Now, is there a way to use this "dataGrid" in the package "lhs" to get

> latin hyper cube samples.

>

> Thanking you

> Prasanna

>

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>

>

Prasanna,

I think I understand your question, please let me know if this explanation is

not what you need. Since, lhs is a contributed package, you could contact me

directly first.

a <- 1:10

b <- 20:30

dataGrid <- expand.grid(a, b)

I believe that you want a Latin hypercube sample from the integers 1-10 in the

first variable and 20-30 in the second. I will offer a way to do something

similar with the lhs package, but then also offer alternatives way which may

meet your needs better.

The lhs package returns a uniformly distributed stratified sample from the

unit hypercube. The marginal distributions can then be transformed to your

distribution of choice. If you wanted a uniform Latin hypercube on [1,10] and

[20,30] with 22 samples, you could do:

require(lhs)

X <- randomLHS(22, 2)

X[,1] <- 1+9*X[,1]

X[,2] <- 20+10*X[,2]

X

OR

X <- randomLHS(22, 2)

X[,1] <- qunif(X[,1], 1, 9)

X[,2] <- qunif(X[,2], 20, 30)

X

Since I think you want integers (which I haven't thought about before now),

then I think we must be careful about what we mean by a Latin hypercube

sample. If you wanted exactly 3 points, then you could divide up the range

[1,10] into three almost equal parts and sample from 1:3, 4:6, and 7:10. The

problem is that it wouldn't be uniform sample across the range. (7 would be

sampled less often than 2 for example)

I think that to do a Latin hypercube sample on the intgers, you should have a

number of integers on the margins which have the number of points sampled as a

common factor. For example if you sample 3 points from 1:9, and 21:32 then

you could sample as follows:

a <- c(sample(1:3,1), sample(4:6, 1), sample(7:9, 1))

b <- c(sample(21:24,1), sample(25:28, 1), sample(29:32,1))

and then randomly permute the entries of a and b.

Or more generally, take n samples from the list of integer groups:

integerLHS <- function(n, intGroups)

{

stopifnot(all(lapply(intGroups, function(X) length(X)%%n)==0))

stopifnot(require(lhs))

stopifnot(is.list(intGroups))

ranges <- lapply(intGroups, function(X) max(X)-min(X))

A <- matrix(nrow=n, ncol=length(intGroups))

for(j in 1:length(ranges))

{

sequ <- order(runif(n))

if(length(intGroups[[1]]) > 1)

{

spacing <- intGroups[[j]][2]-intGroups[[j]][1]

} else stop("must have more than 1 intGroup")

for(k in 1:n)

{

i <- sequ[k]

a <- min(intGroups[[j]])+(i-1)*(ranges[[j]]+spacing)/n

b <- min(intGroups[[j]])+i*(ranges[[j]]+spacing)/n-1

if(a < b)

{

A[k,j] <- sample(seq(a,b,spacing), 1)

} else if(a==b)

{

A[k,j] <- a

} else stop("error")

}

}

return(A)

}

integerLHS(10, list(1:10, 31:40))

integerLHS(5, list(1:10, 31:40))

integerLHS(2, list(1:10, 31:40))

integerLHS(5, list(1:20, 31:60, 101:115))

integerLHS(5, list(seq(2,20,2), 31:60, 101:115))

The function above is neither efficient nor tested, but it is a place for you

to start.

Rob

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