Create a matrix with increment and element with zero subscript
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
<zndeana at 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
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Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/