Hi,
You've got the right idea with make.unique, but you're trying to use repool on
only one population at a time.
First and foremost: don't use row.names(x$tab) or nrow(x$tab), use indNames(x)
and nInd(x), respectively.
Your procedure needs to be done in three steps:
1. sample the individuals
2. rename the individuals
3. repool the populations
Taking the example you gave, this should work:
MySamp8 <- lapply(obj, function(x) x[sample(nInd(x), 8, replace= TRUE)])
MySamp8 <- lapply(MySamp8, function(x){ indNames(x) <-
make.unique(indNames(x)); return(x)})
repool(MySamp8)
Here it is using a custom function:
sample_genind_with_replacement <- function(x, n = 8){
y <- x[sample(nInd(x), n, replace = TRUE)]
indNames(y) <- make.unique(indNames(y))
return(y)
}
MySamp8 <- lapply(obj, sample_genind_with_replacement, n = 8)
repool(MySamp8)
Best,
Zhian
> On Jan 19, 2017, at 07:50 , Bhuller, Ravneet
> <[email protected]> wrote:
>
> Dear Members,
>
> I am working with a genind object which is subset into populations. I want to
> sample 8 individuals randomly with replacement per population and repool them
> in a way that each population has unique individual names.
>
> I am doing the following steps:
>
> obj<- seppop(niger.data.genind)
>
> MySamp8 <- lapply(obj, function(x) x[sample(1:nrow(x$tab), 8, replace= TRUE)])
>
> new_samp8<- lapply(MySamp8, function(x)
> repool(MySamp8[make.unique(row.names(x$tab))]))
>
> But I am getting the following error:
>
> Error in repool(MySamp8[make.unique(row.names(x$tab))]): x is does not
> contain only valid genind objects.
>
> Please if any one can guide me how can I get valid genind objects to repool
> them in a way that each population has unique individual names.
>
> Many thanks for your time,
>
> Rav
>
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>
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