Thanks, Uwe, I think that's it! 
I'll include your fix in the next update.

Regards, Ulrike

---------- Original Message -----------
From: Uwe Ligges <ligges@statistik.tu-dortmund.de> 
To: Ulrike Grömping <groemp@tfh-berlin.de> 
Cc: r-devel@r-project.org 
Sent: Tue, 24 Mar 2009 14:45:57 +0100 
Subject: Re: [Rd]  Error in FrF2 example on Mac OS

> Ulrike Grömping wrote: 
> > Dear all, 
> > 
> > I just noticed that the 0.9 update for FrF2 did not work out for Mac OS due 
> > to an error in an example that ran without error on all other platforms. I 
> > do not find any reason for this. In the past, umlauts or tab characters have 
> > sometimes been an issue, but I didn't find any of these. The function 
> > definition is 
> > 
> > FrF2(nruns = NULL, nfactors = NULL, factor.names = if (!is.null(nfactors)) { 
> > if (nfactors <= 50) Letters[1:nfactors] else 
> > paste("F", 1:nfactors, sep = "")} else NULL, 
> > default.levels = c(-1, 1), generators = NULL, resolution = NULL, 
> > estimable = NULL, max.nfree2fis = FALSE, 
> > randomize = TRUE, seed = NULL, ...){...} 
> > 
> > and the simplest call to this function fails: 
> > FrF2(8,4) 
> > gives the custom error message "nruns must be a power of 2.", which is 
> > generated in the first check within function FrF2: 
> > 
> >     if (!is.null(nruns)){ 
> >        k <- floor(log2(nruns)) 
> >        if (!2^k==nruns) stop("nruns must be a power of 2.")} 
> 
> Probably a rounding issue on different platforms? 
> I guess the test should be something like: 
> 
> if (!is.null(nruns)){ 
>   if(!isTRUE(all.equal(log2(nruns) %% 1, 0))) 
>     stop("nruns must be a power of 2.") 
> } 
> 
> Uwe 
> 
> > Would the Mac (different from all other systems) require FrF2(nruns=8, 
> > nfactors=4) ? Or what else could be the issue here ? 
> > 
> > Thanks for any pointers! 
> > 
> > Regards, Ulrike 
------- End of Original Message -------
 

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