# Latin Hypercube Samples - Questions

## Question 1

I am looking for a package which gives me Latin hypercube samples from a grid of values:

a <- (1:10)
b <- (20:30)
dataGrid <- expand.grid(a, b)

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:

X <- randomLHS(22, 2)
X[,1] <- 1 + 9*X[,1]
X[,2] <- 20 + 10*X[,2]

# OR

Y <- randomLHS(22, 2)
Y[,1] <- qunif(Y[,1], 1, 10)
Y[,2] <- qunif(Y[,2], 20, 30)

#>          [,1]     [,2]
#> [1,] 9.360227 21.96763
#> [2,] 4.808267 28.84202
#> [3,] 1.960651 27.97779
#> [4,] 7.200957 21.79375
#> [5,] 1.200379 28.40089
#> [6,] 4.091125 24.84824
#>          [,1]     [,2]
#> [1,] 8.284968 29.24988
#> [2,] 5.525437 28.53956
#> [3,] 8.981331 27.07328
#> [4,] 2.887880 23.40239
#> [5,] 1.730310 29.07440
#> [6,] 7.708760 27.45095

If you want integers only in the sample, then you can use qinteger.

X <- randomLHS(3, 2)
X[,1] <- qinteger(X[,1], 1, 10)
X[,2] <- qinteger(X[,2], 20, 30)

#>      [,1] [,2]
#> [1,]    4   24
#> [2,]    8   28
#> [3,]    5   20

## Question 2

I am trying to do a Latin hypercube Sampling (LHS) to a 5-parameter design matrix. I want the combination of the first three parameters to sum up to 1 (which obviously do not)

If I divide each of these parameters with the sum, the uniform distribution is lost. Is there a way to maintain the random LHS (with uniformly distributed parameters) so that the referred condition is fulfilled?

In my experience with Latin hypercube samples, most people draw the sample on a uniform hypercube and then transform the uniform cube to have new distributions on the margins. The transformed distributions are not necessarily uniform. It is possible to draw a Latin hypercube with correlated margins and have them sum to one using qdirichlet, but they will not be uniform. I’ll make a quick example argument that explains the difficulty…

In two dimensions, you could draw this which is uniform and correlated.

x <- seq(0.05, 0.95, length = 10)
y <- 1 - x
all.equal(x + y, rep(1, length(x)))
#> [1] TRUE
hist(x, main = "")

hist(y, main = "")

But in three dimensions, it is hard to maintain uniformity because large samples on the first uniform margin overweight the small samples on the other margins.

x <- seq(0.05, 0.95, length = 10)
y <- runif(length(x), 0, 1 - x)
z <- 1 - x - y
hist(x, main = "")

hist(y, main = "")

hist(z, main = "")

The transformed distributions maintain their “Latin” properties, but are in the form of new distributions. The Dirichlet distribution is built for this purpose.

N <- 1000
x <- randomLHS(N, 5)
y <- x
y[,1:3] <- qdirichlet(x[,1:3], c(1, 1, 1))
y[,4] <- x[,4]
y[,5] <- x[,5]

par(mfrow = c(2,3))
dummy <- apply(x, 2, hist, main = "")

par(mfrow = c(2,3))

dummy <- apply(y, 2, hist, main = "")

all.equal(rowSums(y[,1:3]), rep(1, nrow(y)))
#> [1] TRUE

The uniform properties are gone as you can see here…

par(mfrow = c(1,1))
pairs(x)

pairs(y, col = "red")

## Question 3

How do I create a Latin hypercube that ranges between between 0 and 1 and sums to 1?

I have a solution to this problem using a Dirichlet distribution.
The result is not uniformly distributed on (0,1) anymore, but instead is Dirichlet distributed with the parameters alpha. The Latin properties are maintained.

X <- randomLHS(1000, 7)
Y <- qdirichlet(X, rep(1,7))
stopifnot(all(abs(rowSums(Y) - 1) < 1E-12))
range(Y)
#> [1] 5.076098e-06 8.393811e-01

ws <- randomLHS(1000, 7)
wsSums <- rowSums(ws)
wss <- ws / wsSums
stopifnot(all(abs(rowSums(wss) - 1) < 1E-12))
range(wss)
#> [1] 2.648948e-05 5.038924e-01

## Question 5

I need to use Latin hypercube sampling for my own custom functions.

require(lhs)

# functions you described
T1 <- function(t) t*t
WL1 <- function(T1, t) T1*t
BE1 <- function(WL1, T1, t) WL1*T1*t

# t is distributed according to some pdf (e.g. normal)
# draw a lhs with 512 rows and 3 columns (one for each function)
y <- randomLHS(512, 3)
# transform the three columns to a normal distribution (these could be any
# distribution)
t <- apply(y, 2, function(columny) qnorm(columny, 2, 1))
# transform t using the functions provided
result <- cbind(
T1(t[,1]),
WL1(T1(t[,2]), t[,2]),
BE1(WL1(T1(t[,3]), t[,3]), T1(t[,3]), t[,3])
)
# check the results
# these should be approximately uniform
par(mfrow = c(2,2))
dummy <- apply(y, 2, hist, breaks = 50, main = "")
# these should be approximately normal
par(mfrow = c(2,2))

dummy <- apply(t, 2, hist, breaks = 50, main = "")
# these should be the results of the functions
par(mfrow = c(2,2))

dummy <- apply(result, 2, hist, breaks = 50, main = "")

## Question 6

I need a Latin hypercube sample on an integer set or a set of colors.

N <- 1000

x <- randomLHS(N, 4)
y <- as.data.frame(x)
# uniform integers on 1-10
y[,1] <- qinteger(x[,1], 1, 10)
# three colors 1,2,3
y[,2] <- qfactor(x[,2], factor(c("R", "G", "B")))
# other distributions
y[,3] <- qunif(x[,3], 5, 10)
y[,4] <- qnorm(x[,4], 0, 2)

table(y[,1])
#>
#>   1   2   3   4   5   6   7   8   9  10
#> 100 100 100 100 100 100 100 100 100 100
table(y[,2])
#>
#>   B   G   R
#> 334 333 333

hist(y[,3], main="")

hist(y[,4], main="")