It would be helpful to see what your data frames look like ... perhaps you
could share the first few rows of each with us.
dput(head(j1data))
dput(head(predictdata))
Jean
On Mon, Mar 25, 2013 at 9:29 PM, Curtis Burkhalter <
curtisburkhal...@gmail.com> wrote:
> Hello,
>
> I'm working on a problem using nested for-loops and I don't know if it's a
> problem with the order of the loops or something within the loop so any
> help with the problem would be appreciated. To briefly set up the problem.
> I have 259 trees (from 11 different species, of unequal count for each
> species) of which I am trying to predict biomass. For each tree species I
> have 10000 iterations for the regression coefficients, which were estimated
> previously in WinBUGS, and what I'm trying to do is for each tree I want to
> predict the biomass using each species-specific iteration of the regression
> coefficients, so that ultimately each tree has 10000 estimates of biomass,
> organized into a 10000x259 matrix. The input data used in the model
> equation is stored in two separate files and I don't think this is creating
> problems, but I thought it might be worth mentioning. I've pasted the code
> below and if any additional info is needed please write back and I will
> post it.
>
> #read in the model output data from Jenkins eq. 1 under "data"
> #read in the prediction data set under "predict data"
>
> j1data=read.delim("reduced_j1_forR.txt",header=T)
> predictdata=read.delim("predictset_forR.txt",header=T)
> j1data=j1data[,2:4]
> its=c(rep(1:10000,each=11))
> j1data=cbind(its,j1data)
>
> #set up a matrix full of zeros "lnbm" where prediction results are placed
> #set up for loop that first loops over iteration, then species
> #(total iterations=10000it/spp*11spp=110000 iterations)and then over each
> tree
> #(total # of trees=259)
>
> niter=10000
> nspp=11
> ntrees=259
>
> lnbm=matrix(0,10000,259)
> k=numeric()
> for (i in 1:ntrees)
> {
> for (j in 1:nspp)
> {
> for (m in 1:niter)
> {
> k=((j1data$its[m]-1)*1000)+(j1data$spp[j])
> #print(k)
> lnbm[m,i]=j1data$b0[k]+j1data$b1[k]*predictdata$lndbh[i]
> }
> }
> }
>
>
>
> Thanks
>
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>
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