你链接的图片是密度曲线，不是直方图。

如果您一直在阅读ggplot，那么可能您唯一缺少的是将两个数据帧合并为一个长数据帧。

那么，让我们从你拥有的两组独立的数据开始，并将它们结合起来。

carrots <- data.frame(length = rnorm(100000, 6, 2))
cukes <- data.frame(length = rnorm(50000, 7, 2.5))

# Now, combine your two dataframes into one.  
# First make a new column in each that will be 
# a variable to identify where they came from later.
carrots$veg <- 'carrot'
cukes$veg <- 'cuke'

# and combine into your new data frame vegLengths
vegLengths <- rbind(carrots, cukes)

在那之后，如果你的数据已经是长格式的，这是不必要的，你只需要一行来绘制你的图表。

ggplot(vegLengths, aes(length, fill = veg)) + geom_density(alpha = 0.2)

现在，如果你真的想要直方图，下面的方法就可以了。注意，必须从默认的"stack"参数更改位置。如果您不知道您的数据应该是什么样子，您可能会错过这一点。更高的alpha看起来更好。还要注意，我把它做成了密度直方图。很容易删除y = ..density..把它找回来。

ggplot(vegLengths, aes(length, fill = veg)) + 
   geom_histogram(alpha = 0.5, aes(y = ..density..), position = 'identity')

另外，我评论了Dirk的问题，所有的参数都可以简单地在hist命令中。有人问我怎么才能做到。接下来得出的正是德克的数字。

set.seed(42)
hist(rnorm(500,4), col=rgb(0,0,1,1/4), xlim=c(0,10))
hist(rnorm(500,6), col=rgb(1,0,0,1/4), xlim=c(0,10), add = TRUE)

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)。

这里是类似于我只以r为基数给出的ggplot2的版本。我从@nullglob复制了一些。

生成数据

carrots <- rnorm(100000,5,2)
cukes <- rnorm(50000,7,2.5)

您不需要像使用ggplot2那样将其放入数据帧中。这种方法的缺点是你必须写出更多的情节细节。这样做的好处是你可以控制更多的情节细节。

## calculate the density - don't plot yet
densCarrot <- density(carrots)
densCuke <- density(cukes)
## calculate the range of the graph
xlim <- range(densCuke$x,densCarrot$x)
ylim <- range(0,densCuke$y, densCarrot$y)
#pick the colours
carrotCol <- rgb(1,0,0,0.2)
cukeCol <- rgb(0,0,1,0.2)
## plot the carrots and set up most of the plot parameters
plot(densCarrot, xlim = xlim, ylim = ylim, xlab = 'Lengths',
     main = 'Distribution of carrots and cucumbers', 
     panel.first = grid())
#put our density plots in
polygon(densCarrot, density = -1, col = carrotCol)
polygon(densCuke, density = -1, col = cukeCol)
## add a legend in the corner
legend('topleft',c('Carrots','Cucumbers'),
       fill = c(carrotCol, cukeCol), bty = 'n',
       border = NA)

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:

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:

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