Or, of course, if you are willing to reduce it then its just
sum(c) + length(c) * x
On 4/9/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> If c is c(c1, c2, c3) and x is c(x1, x2, x3) then
> c+x is (c1+x1, c2+x2, c3+x3)
> so sum(c+x) is c1+x1+c2+x2+c3+x3 = sum(c) + sum(x)
>
> What you were expecting is given by:
>
> rowSums(outer(1:4, c(-1,0,1), "+")) # gives c(3, 6, 9, 12)
>
> Review the Introduction to R manual and also look at ?outer and ?rowSums
>
>
> On 4/9/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> > Hi, I am writing a function that includes 'sum' function
> > such as:
> > f<-function(x){
> > c<-c(-1,0,1)
> > f<-sum(c+x)
> > }
> > expecting f to be -1+x+0+x+1+x=3x. But I found out that f is
> > sum(x). So, f is always a scalar, which means that f(c(0,1))
> > is not a vector as c(0,3), but 3(0+1)=3. I would like to ask
> > you helping me in solving this problem. I would like to
> > thank you in advance.
> > Sungsu.
> > UC riverside.
> >
> > ______________________________________________
> > [email protected] mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide!
> > http://www.R-project.org/posting-guide.html
> >
>
______________________________________________
[email protected] mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
