[R] nls start values

[Sundar Dorai-Raj]

Wouldn't this be easier?

vals <- 1:100
names(vals) <- sprintf("beta%d", 1:100)
## or
## names(vals) <- paste("beta", 1:100, sep = "")

--sundar

Cal Stats wrote:
> Hi..
>   
>   here is an example
>   
>   ss<-NULL
>   vals<-1:100
>   for(i in 1:100){
>       ss<-c(ss,paste("beta",i,"=",vals[i],sep=""))
>   }
>   sss<-paste(ss,collapse=",")
>   
>   now is there a way i can convert "sss"  so that i can give the command
>   
>   nls(formula,start=sss)
>   
>   Thanks
>   
>   Harsh
> 
> "Charles Annis, P.E." <Charles.Annis at StatisticalEngineering.com> wrote:  Please give us an EXAMPLE of the loop you have in mind.  (It's likely that
> you can use simpler methods than a loop, but without an example we'd be
> guessing.)
> 
> 
> 
> 
> Charles Annis, P.E.
> 
> Charles.Annis at StatisticalEngineering.com
> phone: 561-352-9699
> eFax:  614-455-3265
> http://www.StatisticalEngineering.com
>  
> 
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Cal Stats
> Sent: Saturday, March 11, 2006 3:43 PM
> To: r-help at stat.math.ethz.ch
> Subject: [R] nls start values
> 
> Hi,
>   
>       I have a large number of parameters to estimate in nls say 100:
> beta1--beta100
>   lets say i have 100 values in a vector
>   
>   is there a way where i can create the start vector for nls using a loop
> instead of individually filling the 100 values.
>   
>   Thanks
>   
>   Harsh
>   
>   
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