[R] sum of variables in function

Jonathan P Daily jdaily at usgs.gov
Thu Mar 10 19:34:48 CET 2011


How about:


x <- rnorm(1000)
fn <- function(k) sum(x[1:k]*(x[k+1] + x[k+2]))
vals <- sapply(1:100, fn)
vals

--------------------------------------
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
"Is the room still a room when its empty? Does the room,
 the thing itself have purpose? Or do we, what's the word... imbue it."
     - Jubal Early, Firefly

r-help-bounces at r-project.org wrote on 03/10/2011 08:49:25 AM:

> [image removed] 
> 
> [R] sum of variables in function
> 
> beatleb 
> 
> to:
> 
> r-help
> 
> 03/10/2011 01:15 PM
> 
> Sent by:
> 
> r-help-bounces at r-project.org
> 
> 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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