'a' and 'b' are vectors of length 100, so the sum of them would also
produce a vector of 100. So in one case, your code should look like this:
> a<-replicate(5, rnorm(20))
> b<-replicate(5, rnorm(20))
>
> xyz<-function(x,y){
+ z<-x+y
+ print(paste("the sum of", x, "and", y,"is", z, sep=" "))
+ }
>
> xyz(a,b)
[1] "the sum of 0.078437532748926 and 1.42398359867007 is 1.502421131419"
[2] "the sum of -0.779261070429696 and 0.484730511200338 is
-0.294530559229357"
[3] "the sum of 0.16655966873187 and 0.349236445934426 is
0.515796114666296"
[4] "the sum of 0.265324569774749 and 0.860124249616191 is
1.12544881939094"
[5] "the sum of 0.890780709620268 and 0.404611144608053 is
1.29539185422832"
[6] "the sum of -0.467888369466512 and 0.367044921399527 is
-0.100843448066985"
[7] "the sum of 0.758374556806851 and -1.51919904950798 is
-0.760824492701126"
.........
Is this your expected output? If not, give a clear example of what you are
expecting. Could it be:
> sum(a + b)
[1] 16.90667
On Wed, Jun 19, 2013 at 8:16 PM, Thomas Parr <thomas.p...@maine.edu> wrote:
> I am writing a function and I would like to return the name of a data frame
> in a paste call, but I can't figure out how to just get the name. The names
> of the data frames used, won't be the same each time. I have to be
> overlooking the obvious.
>
> #For example:
>
> a<-replicate(5, rnorm(20))
> b<-replicate(5, rnorm(20))
>
> xyz<-function(x,y){
> z<-x+y
> print(paste("the sum of", x, "and", y,"is", sep=" "))
> z
> }
>
> xyz(a,b)
>
> #That function should return:
>
> >"the sum of a and b is:"
> >.and some data.
>
> Not surprisingly, it returns all the data in the data frame.
>
> Thanks,
> Thomas
>
> ______________________________________________
> R-help@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.
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