Hi Deana,
No, R does not deal with zero subscripts. Could it be done? Of
course. The simplest approach is: 0 + 1 = 1, which is the R
equivalent. You need to adjust your code to go, for example, from 1
to 30 instead of 0 to 29.
Cheers,
Josh
On Mon, Oct 24, 2011 at 4:23 PM, Md Desa, Zairul Nor Deana Binti
<znde...@ku.edu> wrote:
> Hello,
> Does anyone knows how to deal with zero subscript in R. I have this code:
> for (i in 1:nitems){
> + for (j in 1:ncat-1) {
> + draw<-matrix(rnorm(nitems*(ncat-1),seed1,seed2),nitems,(ncat-1))
> + d<-( sigma_d*draw ) + mu_d
> + draw<-matrix(rtnorm((nitems*(ncat-1)),mean = seed1, sd = seed2,
> lower = .1, upper = 1.5),nitems,(ncat-1))
> + d<-(sigma_d*draw) + mu_d
> + write.matrix(cbind(b.i0,d), file = "F:/KU/MIRT
> group/MIMIC-DIF/R/cpprcode/b0d.dat", sep = " ")
> + b[i,j]<-b[i,j-1]+d[i,j]
> + }
> + }
>
> The error as following:
>
> Error in b[i, j ] <- b[i, j - 1] + d[i, j] :
> replacement has length zero
>
> I would like to to have a matrix where the first column takes from initial
> pre-assigned value (b[i,0]+d1), the second column is additive from the first
> column and a constant d2 that is (b(i,1)=b(i,0)+d1+d2, and so forth. Is there
> any way that R can read subscript of zero?
>
> Thanks,
> Deana
>
> ______________________________________________
> 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.
>
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
______________________________________________
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.
