Thanks for your answer.
However, my variable is simulated from the cumulative distribution function of
the Poisson distribution. So, the pattern obtained from the function "qpois" is
not the same as the observed pattern (i.e., obtained from the function "ppois")
set.seed(5)
mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2)
barplot(mortality_probability, names.arg = seq(0, 7, by = 1), xlab = "Age
class", ylab = "Probability")
library(lhs)
set.seed(1)
parm <- c("var1", "var2", "mortality_probability")
X <- randomLHS(100, length(parm))
colnames(X) <- c("var1", "var2", "mortality_probability")
X[, "mortality_probability"] <- qpois(X[, "mortality_probability"], 0.9)
hist(X[, "mortality_probability"])
Thanks for your time
Marine
________________________________
De : S Ellison <s.elli...@lgcgroup.com>
Envoy� : lundi 7 ao�t 2017 14:36
� : Marine Regis; r-help@r-project.org
Objet : RE: Latin hypercube sampling from a non-uniform distribution
> How can I draw a Hypercube sample for the variable mortality_probability so
> that this variable exhibits the same pattern as the observed distribution?
One simple way is to use the uniform random output of randomLHS as input to the
quantile function for your desired distribution(s).
For example:
q <- randomLHS(1000, 3)
colnames(q) <- c("A", "B", "mort")
q[, "mort"] <- qpois(q[,"mort"], 1.5)
S Ellison
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