>>>>> "Bill" == Bill Shipley <[EMAIL PROTECTED]> >>>>> on Wed, 28 Jun 2006 15:53:35 -0400 writes:
Bill> Earlier, I posted the following question: Bill> I want to superimpose histograms from three Bill> populations onto the same graph, changing the shading Bill> of the bars for each population. After consulting the Bill> help files and the archives I cannot find out how to Bill> do this (seemly) simple graph. To be clear, I want Bill> - a single x axis (from -3 to 18) Bill> - three groups of bars forming the histograms of each population Bill> (they will not overlap much, but this is a detail) Bill> - the bars from each histogram having different Bill> shadings or other visually distinguishing features. Bill> Gabor Grothendieck [EMAIL PROTECTED] pointed Bill> to some code to to this but I have found another way Bill> that works even easier. Bill> hist(x[sel1],xlim=c(a,b),ylim=c(A,B)) - this plots the histogram for the Bill> first group (indexed by sel1) but with an x axis and a y axis that spans the Bill> entire range. Bill> par(new=T) - to keep on the same graph Bill> hist(x[sel2],main=Null,xlab=NULL,ylab=NULL,axes=F) -superimposes the second Bill> histogram Bill> par(new=T) - to keep on the same graph Bill> hist(x[sel3],main=Null,xlab=NULL,ylab=NULL,axes=F) -superimposes the third Bill> histogram Hmm, the above does not work (because of 'Null' instead of 'NULL') but even if you fix that, I'm pretty sure you get pretty wrong plots. par(new = TRUE) is quite often a bad choice. In the present case, plot 2 and 3 use different coordinate systems than plot 1. The "correct" solution -- apart from using lattice::histogram() with the correct 'superpose' panel is really to make use of the fact that 1) hist() returns a "histogram" (S3) object which is plotted by 2) plot.histogram() which has a nice help page even though the funcion is namespace-hidden. Here's a reproducible solution --- with quite a bit of extra code in order to show two ``colorizations'' for overplotting, in particular one with ``transparent colors'' : ##MM: construct reproducible example data set.seed(1) x <- c(rnorm(50), (rnorm(60) +3.5), (rnorm(40) +3.5 + 3)) str(grouping <- factor(c(rep(1,50), rep(2,60), rep(3,40)))) (n.gr <- length(table(grouping))) # 3 xr <- range(x) ## Compute all three histograms objects but without plotting: histL <- tapply(x, grouping, hist, plot = FALSE) maxC <- max(print( sapply(lapply(histL, "[[", "counts"), max)) ) # 14 15 15 pdf("3histo.pdf", version = "1.4") # >= 1.4 is needed ## or try the default x11() if((TC <- transparent.cols <- .Device %in% c("pdf", "png"))) { cols <- hcl(h = seq(30, by=360 / n.gr, length = n.gr), l = 65, alpha = 0.5) ## << transparency } else { h.den <- c(10, 15, 20) h.ang <- c(45, 15, -30) } ## Plots the histogram for the ## first group (indexed by sel1) but with an x axis and a y axis that spans the ## entire range. if(TC) { plot(histL[[1]], xlim = xr, ylim= c(0, maxC), col = cols[1], xlab = "x", main = "3 Histograms of 3 Selections from 'x'") } else plot(histL[[1]], xlim = xr, ylim= c(0, maxC), density = h.den[1], angle = h.ang[1], xlab = "x", main = "Histogram of 3 selections from 'x'") ## Using density instead of color make these *add* if(!transparent.cols) { for(j in 2:n.gr) plot(histL[[j]], add = TRUE, density = h.den[j], angle = h.ang[j]) } else { ## use semi-translucent colors (available only for PDF >= 1.4 and PNG devices): for(j in 2:n.gr) plot(histL[[j]], add = TRUE, col = cols[j]) } if(.Device == "pdf") { ## you have set it above dev.off() system("gv 3histo.pdf &")## or use 'acroread' ; 'xpdf' does not show transparent colors... } 1 ##--- Martin Maechler, ETH Zurich ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html