[R] Generating Replicate Datasets (using loops or other means)
Moshe Olshansky
m_olshansky at yahoo.com
Tue Sep 11 02:23:02 CEST 2007
Hi Jonathan,
What exactly do you mean by replication?
Do you want to keep a1,b1,c1,... unchanged but have 30
different sets of random numbers?
Regards,
Moshe.
--- VTLT1999 <jonathan-beard at uiowa.edu> wrote:
>
> Hello All,
>
> I have searched many help forums, message boards,
> etc. and I just can't
> apply the comments to what I need my program to do.
> I am running R 2.5.1 on
> an XP system, and my desire is to produce replicate
> datasets for a
> simulation study I am running. Essentially, I have
> sets of parameters (a's,
> b's, and c's) that define a function which produces
> a decimal value. This
> value is compared to a random uniform value, and is
> coded a 1 if the
> function is greater than the uniform value, 0 if it
> is <= to the uniform
> value. My code thus far works great, but I just
> need it to run several
> times. Here we go:
>
> library(mvtnorm)
> library(sm)
> library(ltm)
> library(irtoys)
>
> k<- 5000
> set.seed(271828)
> t <-
>
rmvnorm(n=k,mean=c(-1,0,1),sigma=matrix(c(1,.8,.5,.8,1,.8,.5,.8,1),3,3))
>
> #Using mv here because of the likely association
> of ability (theta = t)
> across time.
>
> t1<-as.matrix(t[,1])
> t2<-as.matrix(t[,2])
> t3<-as.matrix(t[,3])
>
> set.seed(271828)
>
> # Population item parameters (n=54) from which we
> will select relevant
> items
> # These are the parameters that are used in the
> function
>
> a <- c(1.18120, 0.92613, 0.96886, 0.80503, 1.12384,
> 0.84073, 0.85544, 0.86801, 1.01054, 0.82278,
> 1.10353, 0.78865, 0.98421, 1.76071, 0.89603,
> 0.84671, 0.89737, 0.74775, 0.32190, 0.69730,
> 0.72059, 1.16762, 1.29257, 1.32902, 0.59540,
> 0.51022, 0.59259, 0.93951, 0.68568, 0.55649,
> 0.88084, 0.52940, 0.45735, 0.57560, 1.11779,
> 0.96984, 1.19692, 0.99102, 1.25847, 1.62555,
> 0.63049, 1.07807, 1.04897, 1.23138, 1.14014,
> 1.25230, 1.14844, 0.59287, 0.83143, 0.81723,
> 0.52141, 0.61980, 0.49945, 1.02749)
>
> b <- c(-2.51737, -1.95897, -1.72667, -0.82988,
> -0.36093,
> 0.72554, 0.91442, 0.78061, 0.06088,
> 0.75733,
> -0.76371, 0.24552, -0.42050, 0.88232,
> -0.81761,
> 0.06466, -0.43866, -0.46042, 0.21636,
> -0.73147,
> -1.44086, -1.03718, 0.07275, -0.17197,
> 1.53796,
> -0.45631, -1.69826, -0.66506, 0.98921,
> 0.30714,
> -0.62245, 0.97253, 1.95894, 0.21277,
> 1.96346,
> 1.18825, 1.59917, -0.28401, -1.23530,
> -0.09671,
> -0.31581, -0.66149, -0.81284, -0.35399,
> -0.07623,
> 1.06442, -0.68559, 1.07591, 0.97458,
> 0.06436,
> 1.25622, 1.73954, 1.75052, 2.34088)
>
> c <- c(0.00000, 0.00000, 0.00000, 0.00000, 0.19648,
> 0.31302, 0.26454, 0.19714, 0.06813, 0.21344,
> 0.00000, 0.03371, 0.00000, 0.16581, 0.11054,
> 0.08756, 0.07115, 0.26892, 0.00000, 0.06883,
> 0.00000, 0.14815, 0.32389, 0.19616, 0.17597,
> 0.00000, 0.00000, 0.04337, 0.19949, 0.20377,
> 0.00000, 0.06243, 0.13639, 0.00000, 0.18166,
> 0.15996, 0.20184, 0.08331, 0.24453, 0.26114,
> 0.16434, 0.20750, 0.32658, 0.31870, 0.45227,
> 0.35039, 0.31178, 0.17999, 0.22774, 0.21675,
> 0.10153, 0.17764, 0.15205, 0.19858)
>
> # Item parameters for generating 3PL data for all
> five testing occasions:
> # This selects the relevant parameters for a
> particular data generation run
> # Only parameters for the first testing occasion
> are shown to save space
>
> a1 <- as.matrix(a[c(1:5,15:20,22:24,38:44)])
> b1 <- as.matrix(b[c(1:5,15:20,22:24,38:44)])
> c1 <- as.matrix(c[c(1:5,15:20,22:24,38:44)])
>
> # Here is where I would like to begin my
> replications, but don't know how
> to make R do it.
> # The code below produces a matrix of 0's and 1's
> (which will be used by
> another program)
> # I would like to nest this in a "do loop" such
> that, say, 30 replicate
> datasets are produced using the
> # same parameters.
>
> N <- nrow(t1) # number of examinees
> n <- nrow(a1) # number of items
> d <- 1.7
> theta <- t1
> response <- matrix (0,N,n)
> uni <- matrix (runif(N*n),nrow = N)
>
> for (i in 1:N)
> {
> for (j in 1:n)
> {
> if (
> c1[j]+(1-c1[j])/(1+exp(-d*a1[j]*(theta[i]-b1[j]))) >
> uni[i,j] )
> response[i,j] = 1
> else
> response[i,j] = 0
> }
> }
> write.table(response, file="C:/responses.dat", sep="
> ",row.names=FALSE,
> col.names=FALSE)
>
> I tried earlier nesting this in another for loop,
> but that indexes elements
> of matrices and vectors, and doesn't seem to apply
> to a "global" loop
> methodology. I am attempting to use replicate as we
> speak, but
> documentation is sparse (help("replicate") is nested
> in lapply information).
> Any guidance is greatly appreciated.
>
> Thanks in advance,
>
> Jonathan Beard
>
> --
> View this message in context:
>
http://www.nabble.com/Generating-Replicate-Datasets-%28using-loops-or-other-means%29-tf4418768.html#a12603580
> Sent from the R help mailing list archive at
> Nabble.com.
>
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