R help - Dec 2011 - Summing x1 to x6

Suppose I have the following:

x1<-as.vector(rnorm(10))
x2<-as.vector(rnorm(10))
x3<-as.vector(rnorm(10))
x4<-as.vector(rnorm(10))
x5<-as.vector(rnorm(10))
x6<-as.vector(rnorm(10))
x7<-as.vector(rnorm(10))
x8<-as.vector(rnorm(10))
x9<-as.vector(rnorm(10))
x10<-as.vector(rnorm(10))

I would like the mean of x1 to x6 for each vector position.  I would do
something else with x7-x10.  These vectors are not currently in a dataframe. 
I tried to be clever by trying get(paste(paste("x", 1:6,
sep=""),
collapse="+")) but it didn't work.  Any thoughts are greatly
appreciated,

Pablo

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Not sure what you mean by mean of each position, so here are two
possibilities: rowMeans and colMeans:
> x1<-as.vector(rnorm(10))
> x2<-as.vector(rnorm(10))
> x3<-as.vector(rnorm(10))
> x4<-as.vector(rnorm(10))
> x5<-as.vector(rnorm(10))
> x6<-as.vector(rnorm(10))
> x7<-as.vector(rnorm(10))
> x8<-as.vector(rnorm(10))
> x9<-as.vector(rnorm(10))
> x10<-as.vector(rnorm(10))
>
> # create matrix
> x1.6 <- rbind(x1, x2, x3, x4, x5, x6)
> rowMeans(x1.6)
        x1         x2         x3         x4         x5         x6
-0.7845288  0.3254140 -0.4430038 -0.1533713  0.6074253
-0.1560222
> colMeans(x1.6)
 [1]  0.23480212 -0.08486182 -0.28657007 -0.59151369  0.08995106
0.55138690 -0.22388657
 [8] -0.36354109 -0.20111151 -0.13146655
>

On Mon, Dec 19, 2011 at 11:00 AM, Pablo <pschatfield at gmail.com>
wrote:
> Suppose I have the following:
>
> x1<-as.vector(rnorm(10))
> x2<-as.vector(rnorm(10))
> x3<-as.vector(rnorm(10))
> x4<-as.vector(rnorm(10))
> x5<-as.vector(rnorm(10))
> x6<-as.vector(rnorm(10))
> x7<-as.vector(rnorm(10))
> x8<-as.vector(rnorm(10))
> x9<-as.vector(rnorm(10))
> x10<-as.vector(rnorm(10))
>
> I would like the mean of x1 to x6 for each vector position. ?I would do
> something else with x7-x10. ?These vectors are not currently in a
dataframe.
> I tried to be clever by trying get(paste(paste("x", 1:6,
sep=""),
> collapse="+")) but it didn't work. ?Any thoughts are greatly
appreciated,
>
> Pablo
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/Summing-x1-to-x6-tp4214604p4214604.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

On Dec 19, 2011, at 11:00 AM, Pablo wrote:
> Suppose I have the following:
>
> x1<-as.vector(rnorm(10))
> x2<-as.vector(rnorm(10))
> x3<-as.vector(rnorm(10))
> x4<-as.vector(rnorm(10))
> x5<-as.vector(rnorm(10))
> x6<-as.vector(rnorm(10))
> x7<-as.vector(rnorm(10))
> x8<-as.vector(rnorm(10))
> x9<-as.vector(rnorm(10))
> x10<-as.vector(rnorm(10))
>
> I would like the mean of x1 to x6 for each vector position.
 > apply( data.frame(x1,x2,x3,x4,x5,x6), 1, mean)
  [1] -0.39321854 -0.92940746 -0.04843466 -0.27764111  0.44058599   
0.73512759
  [7] -0.22232332 -0.89535287 -0.33430655  0.66217526
 > apply( matrix(c(x1,x2,x3,x4,x5,x6),ncol=6), 1, mean)
  [1] -0.39321854 -0.92940746 -0.04843466 -0.27764111  0.44058599   
0.73512759
  [7] -0.22232332 -0.89535287 -0.33430655  0.66217526

After seeing Jim Holtman's suggestion, it should be clear that  
rowMeans would work just even better than these `apply` solutions when  
used with this "vertical" orientation of the vectors.

> I would do
> something else with x7-x10.  These vectors are not currently in a  
> dataframe.
> I tried to be clever by trying get(paste(paste("x", 1:6,
sep=""),
> collapse="+")) but it didn't work.  Any thoughts are greatly
> appreciated,
>
-- 

David Winsemius, MD
West Hartford, CT