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)。

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

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)

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)。

Plotly的R API可能对你有用。下图在这里。

library(plotly)
#add username and key
p <- plotly(username="Username", key="API_KEY")
#generate data
x0 = rnorm(500)
x1 = rnorm(500)+1
#arrange your graph
data0 = list(x=x0,
         name = "Carrots",
         type='histogramx',
         opacity = 0.8)

data1 = list(x=x1,
         name = "Cukes",
         type='histogramx',
         opacity = 0.8)
#specify type as 'overlay'
layout <- list(barmode='overlay',
               plot_bgcolor = 'rgba(249,249,251,.85)')  
#format response, and use 'browseURL' to open graph tab in your browser.
response = p$plotly(data0, data1, kwargs=list(layout=layout))

url = response$url
filename = response$filename

browseURL(response$url)

坦白说，我是队里的。