これは、重複するヒストグラムを表すために疑似透明度を使用する、私が作成した関数です。
plotOverlappingHist <- function(a, b, colors=c("white","gray20","gray50"),
breaks=NULL, xlim=NULL, ylim=NULL){
ahist=NULL
bhist=NULL
if(!(is.null(breaks))){
ahist=hist(a,breaks=breaks,plot=F)
bhist=hist(b,breaks=breaks,plot=F)
} else {
ahist=hist(a,plot=F)
bhist=hist(b,plot=F)
dist = ahist$breaks[2]-ahist$breaks[1]
breaks = seq(min(ahist$breaks,bhist$breaks),max(ahist$breaks,bhist$breaks),dist)
ahist=hist(a,breaks=breaks,plot=F)
bhist=hist(b,breaks=breaks,plot=F)
}
if(is.null(xlim)){
xlim = c(min(ahist$breaks,bhist$breaks),max(ahist$breaks,bhist$breaks))
}
if(is.null(ylim)){
ylim = c(0,max(ahist$counts,bhist$counts))
}
overlap = ahist
for(i in 1:length(overlap$counts)){
if(ahist$counts[i] > 0 & bhist$counts[i] > 0){
overlap$counts[i] = min(ahist$counts[i],bhist$counts[i])
} else {
overlap$counts[i] = 0
}
}
plot(ahist, xlim=xlim, ylim=ylim, col=colors[1])
plot(bhist, xlim=xlim, ylim=ylim, col=colors[2], add=T)
plot(overlap, xlim=xlim, ylim=ylim, col=colors[3], add=T)
}
Rの透明色のサポートを使用してそれを行う別の方法があります
a=rnorm(1000, 3, 1)
b=rnorm(1000, 6, 1)
hist(a, xlim=c(0,10), col="red")
hist(b, add=T, col=rgb(0, 1, 0, 0.5) )
結果は次のようになります。