[Rd] can a key of a list be a variable

2009-08-04 Thread Bilel MASMOUDI
Hi,

I search a solution to record data in dynamic structures in R.
I have an algorithm that will be executed each step and whose output is an
array of doubles with unknown size.

The solution I found is to use lists
1) I initialise my list
 l <- list()
2) and at a step numbered i I conacatain the new tab to the list
vect <- algorithm()
l <<--c( l , list(stepi=vect))

The problem is that the key "stepi" is here a constant, I'd like to use a
variable like this
x = paste("step",index,sep="") where index is the last index (i)
l <<--c( l , list(x=vect)) --> this doesn't function, the key is named "x"
not "stepi"
If I write l <<--c( l , list(paste("step",index,sep="") =vect)), I receive
an error

Many thanks for your help
-- 
Bilel

-- 
Bilel

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Re: [Rd] can a key of a list be a variable

2009-08-04 Thread Bilel MASMOUDI
Hi,

I found a solution: creating lists without using keys
l<<-c(l,list(x))

Many thanks,


2009/8/4 Bilel MASMOUDI 

> Hi,
>
> I search a solution to record data in dynamic structures in R.
> I have an algorithm that will be executed each step and whose output is an
> array of doubles with unknown size.
>
> The solution I found is to use lists
> 1) I initialise my list
>  l <- list()
> 2) and at a step numbered i I conacatain the new tab to the list
> vect <- algorithm()
> l <<--c( l , list(stepi=vect))
>
> The problem is that the key "stepi" is here a constant, I'd like to use a
> variable like this
> x = paste("step",index,sep="") where index is the last index (i)
> l <<--c( l , list(x=vect)) --> this doesn't function, the key is named "x"
> not "stepi"
> If I write l <<--c( l , list(paste("step",index,sep="") =vect)), I receive
> an error
>
> Many thanks for your help
> --
> Bilel
>
> --
> Bilel
>



-- 
Bilel

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