Thanks a lot for your help and patience Duncan.
It seems that my questions is really a trivial one but I never realized
that 1:4 means 0 1 2 3 4 , never knew it starts from 0.
On 11 October 2015 at 14:07, Duncan Murdoch <murdoch.dun...@gmail.com>
wrote:
> On 11/10/2015 7:52 AM, Maram SAlem wrote:
> > Dear All,
> > I have a question concerning the seq() function arguments when used in a
> > for() loop.
> > I'll simplify this question to make it more clear. Suppose I have a
> > (5*3)matrix s (for ex.) and I need to write a function with for() loop,
> in
> > each step of the loop I need to generate a sequence whose upper limit is
> > the ith element of the first row of s, then put the resulting sequences
> in
> > a list. I used the following simple code (I've only included the first
> part
> > of the function)
> >
> >
> >> s<-matrix(c(1,0,1,0,1,1,0,0,1,0,0,0,0,0,1),nrow=5,byrow=TRUE)
> >> simpfun<- function (x,n,m,p,alpha,beta)
> > + {
> > + LD<-list()
> > + for (i in 1:m-1)
> > + {
> > + LD[[i]]<-seq(0,x[i],1)
> > + }
> > + print(LD)
> > + }
> >> mk<-simpfun(s[1,],n=6,m=4,p=0.3)
> > Error in seq.default(0, x[i], 1) : 'to' must be of length 1
> >
> > Although x is supposed to be the vector
> > 1 0 1
> > and thus x[1]=1, x[2]=0,x[3]=1.
> > So I don't get why the error "Error in seq.default(0, x[i], 1) : 'to'
> must
> > be of length 1" occurs in the first place.
>
> The range of your loop is 1:m-1, where m is 4. That is
>
> > m <- 4
> > 1:m-1
> [1] 0 1 2 3
>
> and x[0] is length 0.
>
> I think you wanted 1:(m-1) (or even better, seq_len(m-1)) for your loop
> values.
>
> Duncan Murdoch
>
> >
> > Thanks for helping.
> >
> > Maram
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
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> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
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