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
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, JLOne 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
Jean-Louis Abitbol wrote:> 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, MDJean-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 > >
On Sun, 2004-11-21 at 12:31 +0100, 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, JLIt appears that ooplot() is built upon barplot2() to an extent. When I wrote barplot2(), in the case of plotting CI's, it expects that the primary data structure ('data' in ooplot, 'height' in barplot2) have the same dimensions as 'ci.l' and 'ci.u'. Thus, for example: barplot2(dat1[, 2], col = rich.colors(7,"temperature"), names.arg = c("X", "Y", "Z", "A", "B", "C", "D"), plot.ci = TRUE, ci.l = dat2[, 2], ci.u = dat3[, 2], xlab = "Treatment", ylab = "Percent Normalized Patients") will work. Note that I am explicitly passing the requisite data vectors for 'height' and the CI's to the function. In the case of ooplot(), it appears to require that the 'data' argument have at least two columns, which requires that you pass 'dat1' as a two dimensional structure and not dat1[, 2] as a vector. In this case, since ooplot is built upon barplot2, your call to ooplot fails when the check of the 'data' and 'ci.u' dimesional structure takes place. The reason for the failure (even though all three of your data structures appear to be of the same shape initially), is that ooplot does: if (by.row) data <- as.matrix(data) else data <- t(as.matrix(data)) which results in your dat1 being changed from a 7 x 2 matrix to a 2 x 7 matrix. So 'data' now looks like:> dataTT1 TT2 TT3 TT4 TT5 TT6 TT7 1.0 2 3.0 4.0 5.0 6.0 7.0 PointEst 3.6 5 5.8 11.5 7.5 8.7 17.4 ooplot then does: height <- data[-1, , drop = FALSE] to create 'height' which is used later in the function, as it is in barplot2(). So 'height' now looks like:> heightTT1 TT2 TT3 TT4 TT5 TT6 TT7 PointEst 3.6 5 5.8 11.5 7.5 8.7 17.4 The actual checks in the ooplot code (taken verbatim from barplot2) that compare the dimensions of the 'height' argument and the CI's is: if (any(dim(height) != dim(ci.u))) stop("'height' and 'ci.u' must have the same dimensions.") else if (any(dim(height) != dim(ci.l))) stop("'height' and 'ci.l' must have the same dimensions.") Due to the transformations above however, 'height' is now a 1 x 7 matrix, whereas your dat2 and dat3 are 7 x 2 matrices. Hence the failure. So, I suspect that Greg (who I have cc:'d here) needs to look at the ooplot code to make similar transformations on the 'ci.l' and 'ci.u' arguments as he is on the 'data' argument to remove the error. Short term, you have (at least) four options: 1. Use barplot2() as I note above 2. You can modify your call to ooplot(), by using t() on the 'ci.l' and 'ci.u' arguments as follows: ooplot(dat1, type = "barplot", col = rich.colors(7,"temperature"), names.arg = c("X", "Y", "Z", "A", "B", "C", "D"), plot.ci = TRUE, ci.l = t(dat2[, 2]), ci.u = t(dat3[, 2]), xlab = "Treatment", ylab = "Percent Normalized Patients") 3. As Frank has mentioned, you can use his Dotplot() function. 4. Similar to Dotplot() in a fashion, is the plotCI() function, which is also in Greg's gplots package. If you stay with the barplot type of graph, you should consider changing your colors, as the CI's are difficult to discern in the first two columns at least. HTH, Marc Schwartz