Help with ooplot(gplots) and error bars

Dear Pr Harrel,

Thanks for your help at this occasion as well as for previous questions
to the list.

I just looked at the example in your intro doc.

However I am dealing with proportions (ie % of patients responding to a
given treatment).

In this case I am not sure I can use summarize and then the xYplot.

I am not aware of R graphing tools that can deal directly with
proportions adding CI not to mention producing by factor/trellis plots.

This is why I why trying to do it "by hand" (using binconf) with ooplot,
without much success I am afraid. 

Best regards,

JL Abitbol, MD

Jean-Louis,

Here is an example.

# Plot proportions and their Wilson confidence limits
set.seed(3)
d <- expand.grid(continent=c('USA','Europe'), year=1999:2001,
                  reps=1:100)
# Generate binary events from a population probability of 0.2
# of the event, same for all years and continents
d$y <- ifelse(runif(6*100) <= .2, 1, 0)
s <- with(d,
           summarize(y, llist(continent,year),
                     function(y) {
                      n <- sum(!is.na(y))
                      s <- sum(y, na.rm=T)
                      binconf(s, n)
                     }, type='matrix')
)

Dotplot(year ~ Cbind(y) | continent,  data=s, ylab='Year',
         xlab='Probability')

I did have to temporarily override a function in Hmisc to fix a problem. 
  This will be corrected in an upcoming release of Hmisc:

mApply <- function(X, INDEX, FUN=NULL, ..., simplify=TRUE) {
## Matrix tapply
## X: matrix with n rows; INDEX: vector or list of vectors of length n
## FUN: function to operate on submatrices of x by INDEX
## ...: arguments to FUN; simplify: see sapply
## Modification of code by Tony Plate <tplate at blackmesacapital.com> 10Oct02
## If FUN returns more than one number, mApply returns a matrix with
## rows corresponding to unique values of INDEX
nr <- nrow(X)
if(!length(nr)) {  ## X not a matrix
   r <- tapply(X, INDEX, FUN, ..., simplify=simplify)
   if(is.matrix(r)) r <- drop(t(r)) else
   if(simplify && is.list(r))
     r <- drop(matrix(unlist(r), nrow=length(r),
                      dimnames=list(names(r),names(r[[1]])), byrow=TRUE))
} else {
   idx.list <- tapply(1:nr, INDEX, c)
   r <- sapply(idx.list, function(idx,x,fun,...) 
fun(x[idx,,drop=FALSE],...),
               x=X, fun=FUN, ..., simplify=simplify)
   if(simplify) r <- drop(t(r))
}
dn <- dimnames(r)
if(length(dn) && !length(dn[[length(dn)]])) {
   fx <- FUN(X,...)
   dnl <- if(length(names(fx))) names(fx) else dimnames(fx)[[2]]
   dn[[length(dn)]] <- dnl
   dimnames(r) <- dn
}

if(simplify && is.list(r) && is.array(r)) {

   ll <- sapply(r, length)
   maxl <- max(ll)
   empty <- (1:length(ll))[ll==0]
   for(i in empty) r[[i]] <- rep(NA, maxl)
   ## unlist not keep place for NULL entries for nonexistent categories
   first.not.empty <- ((1:length(ll))[ll > 0])[1]
   nam <- names(r[[first.not.empty]])
   dr <- dim(r)

   r <- aperm(array(unlist(r), dim=c(maxl,dr),
                    dimnames=c(list(nam),dimnames(r))),
              c(1+seq(length(dr)), 1))
}
r
}



Frank

On Sun, 21 Nov 2004 07:48:58 -0500, "Frank E Harrell Jr"
<f.harrell at vanderbilt.edu> said:

Jean-Louis Abitbol wrote:

Dear All

I am trying to graph a proportion and CI95% by a factor with ooplot (any
other better solution ?)

It works well until I try to add the confidence interval.

this is the error message and and a description of the data:
 

dat1

       PointEst
TT1   1      3.6
TT2   2      5.0
TT3   3      5.8
TT4   4     11.5
TT5   5      7.5
TT5   6      8.7
TT7   7     17.4

dat2

       Lower
TT1   1   1.0
TT2   2   2.2
TT3   3   2.7
TT4   4   6.7
TT5   5   3.9
TT5   6   4.6
TT7   7  11.5

dat3

       Upper
TT1   1  12.3
TT2   2  11.2
TT3   3  12.1
TT4   4  19.1
TT5   5  14.2
TT5   6  15.6
TT7   7  25.6

ooplot(dat1,type="barplot",col=rich.colors(7,"temperature"),names.arg=c("X","Y","Z","A","B","C","D"),plot.ci=T,

+ ci.l=dat2,ci.u=dat3, xlab="Treatment", ylab="Percent Normalized
Patients")
Error in ooplot.default(dat1, type = "barplot", col = rich.colors(7,
"temperature"),  : 
       'height' and 'ci.u' must have the same dimensions.

I have tried various ways of supplying ci.l and ci.u (including a
vector)


Thanks for the help that anyone can bring,

Regards, JL

One way is to look at the examples for Dotplot in the Hmisc package. 
Those examples display bootstrap percentile confidence intervals for a
mean.

-- 
Frank E Harrell Jr   Professor and Chair           School of Medicine
                     Department of Biostatistics   Vanderbilt University