Hi:
Assuming that z constitutes the last two elements of your input vector,
try2 <- function(x) {
n <- length(x)
y <- x[1:(n - 2)]
z <- x[n - 1] + x[n]
sum(y) * z
}
x0 <- 1:4 # result should be 21
x1 <- 1:8 # z = 15, result = 21 * 15 = 315
try2(x0)
try2(x1)
HTH,
Dennis
On Thu, Mar 10, 2011 at 5:49 AM, beatleb <rhelpfo...@gmail.com> wrote:
> Dear R users,
>
> Probably, this is quite a simpe question, but I do not find the proper way
> to obtain want I need. To explain the problem, I constructed a simple
> example.
>
> Suppose I have the following function:
>
> try1<-function(x){
> y<-x[1:2]
> z<-x[3:4]
> y[1]*(z[1]+z[2])+y[2]*(z[1]+z[2])
> }
>
> This function will be part of a for loop. This is what I like to obtain for
> every k:
>
> if k=2
> try1<-function(x){
> y<-x[1:2]
> z<-x[3:4]
> y[1]*(z[1]+z[2])+y[2]*(z[1]+z[2])
> }
>
> if k=3
> try1<-function(x){
> y<-x[1:3]
> z<-x[4:5]
> y[1]*(z[1]+z[2])+y[2]*(z[1]+z[2])+y[3]*(z[1]+z[2])
> }
>
> if k=4
> try1<-function(x){
> y<-x[1:3]
> z<-x[4:5]
> y[1]*(z[1]+z[2])+y[2]*(z[1]+z[2])+y[3]*(z[1]+z[2])+y[4]*(z[1]+z[2])
> }
>
> Assume that k will be quite high, for example 100.
> The problem isn't in defining a for loop, or in defining the appropriate y
> or z, but in the line starting with
> y[1]. The values of the variables are still unknown, which causes that I am
> not able to create this sum.
>
> I really hope that there is somebody in the audience that can help me.
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/sum-of-variables-in-function-tp3345888p3345888.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
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> PLEASE do read the posting guide
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>
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