CRAN Package Check Results for Package pan

Last updated on 2020-08-07 01:49:54 CEST.

Check Details

Version: 1.6
Check: examples
Result: ERROR
    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)
    > y6 <- result$y
    > result <- pan(y,subj,pred,xcol,zcol,prior,seed=9029,iter=100,start=result$last)
    > y7 <- result$y
    > result <- pan(y,subj,pred,xcol,zcol,prior,seed=4343,iter=100,start=result$last)
    > y8 <- result$y
    > result <- pan(y,subj,pred,xcol,zcol,prior,seed=2372,iter=100,start=result$last)
    > y9 <- result$y
    > result <- pan(y,subj,pred,xcol,zcol,prior,seed=7081,iter=100,start=result$last)
    > y10 <- result$y
    > ########################################################################
    > # Now we combine the imputation results according to mitools
    > ########################################################################
    > # First, we build data frames from above,
    > d1 <- data.frame(y=y1,subj,pred)
    > d2 <- data.frame(y=y2,subj,pred)
    > d3 <- data.frame(y=y3,subj,pred)
    > d4 <- data.frame(y=y4,subj,pred)
    > d5 <- data.frame(y=y5,subj,pred)
    > d6 <- data.frame(y=y6,subj,pred)
    > d7 <- data.frame(y=y7,subj,pred)
    > d8 <- data.frame(y=y8,subj,pred)
    > d9 <- data.frame(y=y9,subj,pred)
    > d10 <- data.frame(y=y10,subj,pred)
    > # Second, we establish a S3 object as needed for the function MIcombine
    > # nevertheless we start with an ordinary least squares regression
    > require(mitools)
    Loading required package: mitools
    > d <- imputationList(list(d1,d2,d3,d4,d5,d6,d7,d8,d9,d10))
    > w <- with(d,lm(y~-1+pred))
    > MIcombine(w)
    Multiple imputation results:
     with(d, lm(y ~ -1 + pred))
     MIcombine.default(w)
     results se
    predint -3.367055 4.126646
    preddummy1 11.446035 5.439205
    preddummy2 20.255944 5.495544
    preddummy3 20.305059 5.633694
    preddummy4 4.546684 5.933009
    preddummy5 10.922610 5.495544
    > # Now, we can turn to lmer as in lme4 package but in this case it is the
    > # same.
    > if(require(lme4)) {
    + w2 <- with(d,lmer(y~-1+pred+(1|subj)))
    + b <- MIextract(w2,fun=fixef)
    + Var <- function(obj) unlist(lapply(diag(vcov(obj)),function(m) m))
    + v <- MIextract(w2,fun=Var)
    + MIcombine(b,v)
    + detach(marijuana)
    + }
    Loading required package: lme4
    Loading required package: Matrix
    Error in is.nloptr(ret) : objective in x0 returns NA
    Calls: with ... withCallingHandlers -> do.call -> <Anonymous> -> nloptr -> is.nloptr
    Execution halted
Flavor: r-patched-solaris-x86