这么多伟大的答案，但由于我刚刚写了一个函数('basicPlotteR'包中的plotMultipleHistograms())函数来做到这一点，我想我会添加另一个答案。

这个函数的优点是，它自动设置适当的X轴和Y轴限制，并定义在所有分布中使用的公共容器集。

下面是如何使用它:

# Install the plotteR package
install.packages("devtools")
devtools::install_github("JosephCrispell/basicPlotteR")
library(basicPlotteR)

# Set the seed
set.seed(254534)

# Create random samples from a normal distribution
distributions <- list(rnorm(500, mean=5, sd=0.5), 
                      rnorm(500, mean=8, sd=5), 
                      rnorm(500, mean=20, sd=2))

# Plot overlapping histograms
plotMultipleHistograms(distributions, nBins=20, 
                       colours=c(rgb(1,0,0, 0.5), rgb(0,0,1, 0.5), rgb(0,1,0, 0.5)), 
                       las=1, main="Samples from normal distribution", xlab="Value")

plotMultipleHistograms()函数可以接受任意数量的分布，并且所有一般的绘图参数都应该与它一起工作(例如:las, main等)。

下面是一个更简单的解决方案，使用基本图形和alpha混合(并不适用于所有图形设备):

set.seed(42)
p1 <- hist(rnorm(500,4))                     # centered at 4
p2 <- hist(rnorm(500,6))                     # centered at 6
plot( p1, col=rgb(0,0,1,1/4), xlim=c(0,10))  # first histogram
plot( p2, col=rgb(1,0,0,1/4), xlim=c(0,10), add=T)  # second

关键是颜色是半透明的。

编辑，两年多后:由于这篇文章刚刚获得了好评，我想我不妨添加一个可视化的代码，因为alpha混合是如此有用:

下面是我写的一个函数，它使用伪透明来表示重叠的直方图

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)
}

下面是使用R对透明颜色的支持的另一种方法

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) )

最终的结果是这样的:

已经有了漂亮的答案，但我想加上这个。我觉得不错。 (从@Dirk复制随机数)。需要图书馆(天平)

set.seed(42)
hist(rnorm(500,4),xlim=c(0,10),col='skyblue',border=F)
hist(rnorm(500,6),add=T,col=scales::alpha('red',.5),border=F)

结果是……

更新:这个重叠函数可能对某些人也有用。

hist0 <- function(...,col='skyblue',border=T) hist(...,col=col,border=border) 

我觉得过去的结果比过去的好看

hist2 <- function(var1, var2,name1='',name2='',
              breaks = min(max(length(var1), length(var2)),20), 
              main0 = "", alpha0 = 0.5,grey=0,border=F,...) {    

library(scales)
  colh <- c(rgb(0, 1, 0, alpha0), rgb(1, 0, 0, alpha0))
  if(grey) colh <- c(alpha(grey(0.1,alpha0)), alpha(grey(0.9,alpha0)))

  max0 = max(var1, var2)
  min0 = min(var1, var2)

  den1_max <- hist(var1, breaks = breaks, plot = F)$density %>% max
  den2_max <- hist(var2, breaks = breaks, plot = F)$density %>% max
  den_max <- max(den2_max, den1_max)*1.2
  var1 %>% hist0(xlim = c(min0 , max0) , breaks = breaks,
                 freq = F, col = colh[1], ylim = c(0, den_max), main = main0,border=border,...)
  var2 %>% hist0(xlim = c(min0 , max0),  breaks = breaks,
                 freq = F, col = colh[2], ylim = c(0, den_max), add = T,border=border,...)
  legend(min0,den_max, legend = c(
    ifelse(nchar(name1)==0,substitute(var1) %>% deparse,name1),
    ifelse(nchar(name2)==0,substitute(var2) %>% deparse,name2),
    "Overlap"), fill = c('white','white', colh[1]), bty = "n", cex=1,ncol=3)

  legend(min0,den_max, legend = c(
    ifelse(nchar(name1)==0,substitute(var1) %>% deparse,name1),
    ifelse(nchar(name2)==0,substitute(var2) %>% deparse,name2),
    "Overlap"), fill = c(colh, colh[2]), bty = "n", cex=1,ncol=3) }

的结果

par(mar=c(3, 4, 3, 2) + 0.1) 
set.seed(100) 
hist2(rnorm(10000,2),rnorm(10000,3),breaks = 50)

is

基本想法很好，但代码可以改进。[需要很长时间解释，因此需要单独回答，而不是评论。]

hist()函数默认绘制图形，因此需要添加plot=FALSE选项。此外，通过plot(0,0,type="n"，…)调用可以更清晰地建立plot区域，您可以在其中添加轴标签、plot标题等。最后，我想提一下，还可以使用阴影来区分两个直方图。代码如下:

set.seed(42)
p1 <- hist(rnorm(500,4),plot=FALSE)
p2 <- hist(rnorm(500,6),plot=FALSE)
plot(0,0,type="n",xlim=c(0,10),ylim=c(0,100),xlab="x",ylab="freq",main="Two histograms")
plot(p1,col="green",density=10,angle=135,add=TRUE)
plot(p2,col="blue",density=10,angle=45,add=TRUE)

这里是结果(有点太宽，因为RStudio:-)):

下面是一个在“经典”R图形中如何做到这一点的例子:

## generate some random data
carrotLengths <- rnorm(1000,15,5)
cucumberLengths <- rnorm(200,20,7)
## calculate the histograms - don't plot yet
histCarrot <- hist(carrotLengths,plot = FALSE)
histCucumber <- hist(cucumberLengths,plot = FALSE)
## calculate the range of the graph
xlim <- range(histCucumber$breaks,histCarrot$breaks)
ylim <- range(0,histCucumber$density,
              histCarrot$density)
## plot the first graph
plot(histCarrot,xlim = xlim, ylim = ylim,
     col = rgb(1,0,0,0.4),xlab = 'Lengths',
     freq = FALSE, ## relative, not absolute frequency
     main = 'Distribution of carrots and cucumbers')
## plot the second graph on top of this
opar <- par(new = FALSE)
plot(histCucumber,xlim = xlim, ylim = ylim,
     xaxt = 'n', yaxt = 'n', ## don't add axes
     col = rgb(0,0,1,0.4), add = TRUE,
     freq = FALSE) ## relative, not absolute frequency
## add a legend in the corner
legend('topleft',c('Carrots','Cucumbers'),
       fill = rgb(1:0,0,0:1,0.4), bty = 'n',
       border = NA)
par(opar)

唯一的问题是，如果直方图断点是对齐的，它看起来会更好，这可能需要手动完成(在传递给hist的参数中)。