Running examples in ‘pan-Ex.R’ failed
The error most likely occurred in:

> ### Name: pan
> ### Title: Imputation of multivariate panel or cluster data
> ### Aliases: pan
> ### Keywords: models
> 
> ### ** Examples
> 
> ########################################################################
> # This example is somewhat atypical because the data consist of a
> # single response variable (change in heart rate) measured repeatedly;
> # most uses of pan() will involve r > 1 response variables. If we had
> # r response variables rather than one, the only difference would be
> # that the vector y below would become a matrix with r columns, one
> # for each response variable. The dimensions of Sigma (the residual
> # covariance matrix for the response) and Psi (the covariance matrix
> # for the random effects) would also change to (r x r) and (r*q x r*q),
> # respectively, where q is the number of random coefficients in the
> # model (in this case q=1 because we have only random intercepts). The
> # new dimensions for Sigma and Psi will be reflected in the prior
> # distribution, as Dinv and Binv become (r x r) and (r*q x r*q).  
> #
> # The pred matrix has the same number of rows as y, the number of
> # subject-occasions. Each column of Xi and Zi must be represented in
> # pred. Because Zi is merely the first column of Xi, we do not need to
> # enter that column twice. So pred is simply the matrix Xi, stacked
> # upon itself nine times.
> #
> data(marijuana)
> attach(marijuana)
> pred <- with(marijuana,cbind(int,dummy1,dummy2,dummy3,dummy4,dummy5))
> #
> # Now we must tell pan that all six columns of pred are to be used in
> # Xi, but only the first column of pred appears in Zi.
> #
> xcol <- 1:6
> zcol <- 1
> ########################################################################
> # The model specification is now complete. The only task that remains
> # is to specify the prior distributions for the covariance matrices
> # Sigma and Psi.
> #
> # Recall that the dimension of Sigma is (r x r) where r
> # is the number of response variables (in this case, r=1). The prior
> # distribution for Sigma is inverted Wishart with hyperparameters a 
> # (scalar) and Binv (r x r), where a is the imaginary degrees of freedom
> # and Binv/a is the prior guesstimate of Sigma. The value of a must be
> # greater than or equal to r. The "least informative" prior possible
> # would have a=r, so here we will take a=1. As a prior guesstimate of 
> # Sigma we will use the (r x r) identity matrix, so Binv = 1*1 = 1.
> #
> # By similar reasoning we choose the prior distribution for Psi. The
> # dimension of Psi is (r*q x r*q) where q is the number of random
> # effects in the model (i.e. the length of zcol, which in this case is
> # one). The hyperparameters for Psi are c and Dinv, where c is the
> # imaginary degrees of freedom (which must be greater than or equal to
> # r*q) and Dinv/c is the prior guesstimate of Psi. We will take c=1
> # and Dinv=1*1 = 1.
> #
> # The prior is specified as a list with four components named a, Binv,
> # c, and Dinv, respectively.
> #
> prior <- list(a=1,Binv=1,c=1,Dinv=1)
> ########################################################################
> # Now we are ready to run pan(). Let's assume that the pan function
> # and the object code have already been loaded into R. First we
> # do a preliminary run of 1000 iterations. 
> #
> result <- pan(y,subj,pred,xcol,zcol,prior,seed=13579,iter=1000)
> #
> # Check the convergence behavior by making time-series plots and acfs
> # for the model parameters. Variances will be plotted on a log
> # scale. We'll assume that a graphics device has already been opened.
> #
> plot(1:1000,log(result$sigma[1,1,]),type="l")
> acf(log(result$sigma[1,1,]))
> plot(1:1000,log(result$psi[1,1,]),type="l")
> acf(log(result$psi[1,1,]))
> par(mfrow=c(3,2))
> for(i in 1:6) plot(1:1000,result$beta[i,1,],type="l")
> for(i in 1:6) acf(result$beta[i,1,])
> #
> # This example appears to converge very rapidly; the only appreciable
> # autocorrelations are found in Psi, and even those die down by lag
> # 10. With a sample this small we can afford to be cautious, so let's
> # impute the missing data m=10 times taking 100 steps between
> # imputations. We'll use the current simulated value of y as the first
> # imputation, then restart the chain where we left off to produce
> # the second through the tenth.
> #
> y1 <- result$y
> result <- pan(y,subj,pred,xcol,zcol,prior,seed=9565,iter=100,start=result$last)
> y2 <- result$y
> result <- pan(y,subj,pred,xcol,zcol,prior,seed=6047,iter=100,start=result$last)
> y3 <- result$y
> result <- pan(y,subj,pred,xcol,zcol,prior,seed=3955,iter=100,start=result$last)
> y4 <- result$y
> result <- pan(y,subj,pred,xcol,zcol,prior,seed=4761,iter=100,start=result$last)
> y5 <- result$y
> result <- pan(y,subj,pred,xcol,zcol,prior,seed=9188,iter=100,start=result$last)
Error in pan(y, subj, pred, xcol, zcol, prior, seed = 9188, iter = 100,  : 
  NA/NaN/Inf in foreign function call (arg 20)
Execution halted