# [R] Latin Hypercube Sampling when parameters are defined according to specific probability distributions

Bert Gunter bgunter.4567 at gmail.com
Thu Jun 1 16:48:54 CEST 2017

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Bert

Bert Gunter

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On Wed, May 31, 2017 at 9:49 PM, Nelly Reduan <nell.redu at hotmail.fr> wrote:
> Thank you very much Rob for your answer. I have some difficulties to understand how to apply my agent-based model to each parameter combination generated by the LHS, in particular when parameters are defined by probability distributions. Indeed, I have multiple parameters in my model: parameters which are defined by a single value (like “temperature", "pressure”) and parameters which are defined by probability distributions (like “dispersal distance”). It’s correct for me to treat distance as a class. When all parameters are defined by a single value, I need first to create a data frame in which each column represents a different parameter, and each row represents a different combination of parameter values. Then, I apply my model to each row of the data frame. But, it’s not clear for me how to do this when parameters are defined from probability distributions? In particular, how can I use your code to apply my model to each of the 50 rows of the data frame “tabLHS”? Given that one row corresponds to one model simulation, I should have a value generated by the LHS for all distance classes at the first line of the data frame.
>
>
>
> library(pse)
> q <- list("qexp", "qunif", "qunif")
> q.arg <- list(list(rate=exponential_rate), list(min=0, max=1),
> list(min=0, max=1))
> uncoupledLHS <- LHS(model=model_function, input_parameters, N, q, q.arg)
> hist(uncoupledLHS\$data\$dispersal_distance, breaks=10)
>
> tabLHS <- get.data(uncoupledLHS)
>
>
>
> Sorry, it’s the first time that I perform a sensitivity analysis using the LHS.
>
>
> Thank you very much for your time.
>
> Have a nice day
>
> Nell
>
>
> ________________________________
> De : Rob C <bertcarnell at gmail.com>
> Envoyé : mardi 30 mai 2017 16:26:08
> À : Nelly Reduan
> Cc : r-help at r-project.org
> Objet : Re: [R] Latin Hypercube Sampling when parameters are defined according to specific probability distributions
>
> Nell,
>
> I still might not be interpreting your question correctly, but I will try.
>
> When you say that the sum of the probabilities of all distance classes
> must equal one, I am going to treat distance as a class, instead of as a
> continuous variable.
>
> distance_class_probabilities <- c(9, 11, 10, 8.9, 8, 7, 6, 5.8, 5.1,
> 4, 3.9, 3.7, 3.4, 3.1, 2, 1.9, 1.6, 1.4, 1, 0.9, 0.8, 0.7, 0.4, 0.3,
> 0.1)/100
> sum(distance_class_probabilities)
> distance_classes <- factor(paste0("d", 1:25), ordered = TRUE,
> levels=paste0("d", 1:25))
> input_parameters <- c("dispersal_distance", "temperature", "pressure")
> N <- 1000
>
> plot(1:25, distance_class_probabilities, type="h", lwd=5)
>
> set.seed(1)
> require(lhs)
> X <- randomLHS(N, length(input_parameters))
> dimnames(X) <- list(NULL, input_parameters)
> Y <- X
> Y[,"dispersal_distance"] <-
> approx(x=cumsum(distance_class_probabilities), y=1:25,
> xout=X[,"dispersal_distance"], method="constant", yleft=0)\$y + 1
>
> hist(Y[,"dispersal_distance"], breaks=c(seq(0.5, 25.5, by=1)))
> plot(table(distance_classes[Y[,"dispersal_distance"]]))
>
>
> Is it still a Latin hypercube?
>
> This is a more difficult question.  In some ways, the sample is still a Latin
> hypercube since it was drawn that way.  But once the sample has been discretized
> into the distance classes, then it loses the latin property of having only one
> sample per "row".  It might be close enough for your purposes though.
>
> Rob
>
>
>
> On Tue, May 30, 2017 at 10:59 AM, Nelly Reduan <nell.redu at hotmail.fr> wrote:
>> parameter “dispersal distance”. The sum of the probabilities of all distance
>> classes must be equal to 1:
>>
>> y <- c(9, 11, 10, 8.9, 8, 7, 6, 5.8, 5.1, 4, 3.9, 3.7, 3.4, 3.1, 2, 1.9,
>> 1.6, 1.4, 1, 0.9, 0.8, 0.7, 0.4, 0.3, 0.1)
>>
>> x <- seq(1, 25, by = 1)
>>
>> barplot(y/100, names.arg=x, ylab="Probability", xlab="Distance (km)")
>>
>>
>>
>> With this condition, is it possible to perform a LHS?
>>
>> Thanks a lot for your time.
>>
>> Nell
>>
>>
>> ________________________________
>> De : R-help <r-help-bounces at r-project.org> de la part de Rob C
>> <bertcarnell at gmail.com>
>> Envoyé : samedi 27 mai 2017 13:32:23
>> À : r-help at r-project.org
>> Objet : Re: [R] Latin Hypercube Sampling when parameters are defined
>> according to specific probability distributions
>>
>>>May 26, 2017; 11:41am  Nelly Reduan Latin Hypercube Sampling when
>>> parameters are >defined according to specific probability distributions
>>>Hello,
>>> I would like to perform a sensitivity analysis using a Latin Hypercube
>>> Sampling (LHS).
>>>Among the input parameters in the model, I have a parameter dispersal
>>> distance which is defined according to an exponential probability
>>> distribution.
>>
>>>In the model, the user thus sets a default probability value for each
>>> distance class.
>>
>>>For example, for distances ([0  2]; ]2  4]; ]4  6]; ]6  8]; ]8  10];; ]48
>>> 50],
>>
>>>respective probabilities are 0.055; 0.090; 0.065; 0.035; 0.045;; 0.005.
>>
>>  >Here is the code to represent an exponential probability
>> distribution for the parameter dispersal distance:
>>
>>>set.seed(0)
>>>foo <- rexp(100, rate = 1/10)
>>>hist(foo, prob=TRUE, breaks=20, ylim=c(0,0.1), xlab ="Distance (km)")
>>>lines(dexp(seq(1, 100, by = 1), rate = 1/mean(foo)),col="red")
>>>1/mean(foo)
>>
>>>When a parameter is defined according to a specific probability
>>> distribution, how can I perform a LHS ?
>>>For example, should I sample N values from a uniform distribution for each
>>> distance class (i.e., [0 � 2]; ]2 � 4]; ]4 � 6]; ]6 � 8]; ]8 � 10];��; ]48 �
>>> 50])
>>>or sample N values from exponential distributions with different rates ?
>>
>>>Here is the code used to perform a LHS when the parameter �dispersal
>>> distance� is defined by one default value in the model:
>>
>>>library(pse)
>>>factors <- c("distance")
>>>q <- c("qexp")
>>>q.arg <- list( list(rate=1/30) )
>>>uncoupledLHS <- LHS(model=NULL, factors, 50, q, q.arg)
>>
>>>Thanks a lot for your time.
>>>Have a nice day
>>>Nell
>>
>> Nell,
>>
>> I would like to suggest a slightly different method for generating the
>> sample using the lhs library,  then I will try using the pse library.
>> Generally when you have a package specific
>> question, you should try to contact the package maintainer first.
>>
>> set.seed(1)
>> # I don't think your model has only one parameter, so I will include
>> multiple
>> input_parameters <- c("dispersal_distance", "temperature", "pressure")
>> N <- 50
>> exponential_rate <- 1/30
>>
>> library(lhs)
>> X <- randomLHS(N, length(input_parameters))
>> dimnames(X) <- list(NULL, input_parameters)
>> # X is now a uniformly distributed Latin hypercube
>> hist(X[,1], breaks=5)
>> hist(X[,2], breaks=5)
>> hist(X[,3], breaks=5)
>> # now, transform the dispersal_distance paramter to an exponential sample
>> Y <- X
>> Y[,"dispersal_distance"] <- qexp(X[,"dispersal_distance"],
>> rate=exponential_rate)
>> hist(Y[,1], breaks=10)
>> # you can transform the other marginals as required and then assess
>> function sensitivity
>> model_function <- function(z) z[1]*z[2] + z[3]
>> apply(Y, 1, model_function)
>>
>> # now, trying to use pse
>> library(pse)
>> q <- list("qexp", "qunif", "qunif")
>> q.arg <- list(list(rate=exponential_rate), list(min=0, max=1),
>> list(min=0, max=1))
>> uncoupledLHS <- LHS(model=model_function, input_parameters, N, q, q.arg)
>> hist(uncoupledLHS\$data\$dispersal_distance, breaks=10)
>>
>> Rob
>>
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>> and provide commented, minimal, self-contained, reproducible code.
>
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