# [R] Summing x1 to x6

David Winsemius dwinsemius at comcast.net
Mon Dec 19 20:30:14 CET 2011

```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)
 -0.39321854 -0.92940746 -0.04843466 -0.27764111  0.44058599
0.73512759
 -0.22232332 -0.89535287 -0.33430655  0.66217526
> apply( matrix(c(x1,x2,x3,x4,x5,x6),ncol=6), 1, mean)
 -0.39321854 -0.92940746 -0.04843466 -0.27764111  0.44058599
0.73512759
 -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

```