R help - Aug 2017 - Latin hypercube sampling from a non-uniform distribution

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.Ellison at LGCGroup.com>
Envoy? : lundi 7 ao?t 2017 14:36
? : Marine Regis; r-help at 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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> However, my variable is simulated from the cumulative distribution function
> of the Poisson distribution. 
Then I am afraid I don't know what you're trying to achieve. 
Or why.

However, the principle holds; write a function that maps [0,1] to the
'pattern' you want, do that and apply it to the result from randomLHS.
It happens that for generating data that follow a given probability distribution
F, that function is the quantile function for F so you often do not need to
write it.

 




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This email and any attachments are confidential. Any use...{{dropped:8}}

Thanks for your answer.

I have attached the plot for representing the variable. I think that I need to
draw a Hypercube sample for each age class (i.e., for 0, 1, 2, 3, 4, 5, 6, 7) in
a given simulation (i.e., N = 1) and the LHS values for all age classes should
be like the observed cumulative distribution (see attached figure). Thus, the
output of randomLHS should be a matrix with 100 rows (N = 100 simulations) and 7
columns (7 age classes) containing LHS values and each row should exhibit the
same pattern as the observed cumulative distribution.
With the command ?qpois(X[, "mortality_probability"], 0.9)?, I don?t
obtain a LHS value for each age class and the distribution is not a cumulative
distribution as in the attached figure. So, I am afraid I don?t know how to do
that.
Thanks so much for your time
Marine



________________________________
De : S Ellison <S.Ellison at LGCGroup.com>
Envoy? : mardi 8 ao?t 2017 14:48
? : Marine Regis; r-help at r-project.org
Objet : RE: Latin hypercube sampling from a non-uniform distribution
> However, my variable is simulated from the cumulative distribution function
> of the Poisson distribution.
Then I am afraid I don't know what you're trying to achieve.
Or why.

However, the principle holds; write a function that maps [0,1] to the
'pattern' you want, do that and apply it to the result from randomLHS.
It happens that for generating data that follow a given probability distribution
F, that function is the quantile function for F so you often do not need to
write it.






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