我正在使用 R,我有两个数据框:胡萝卜和黄瓜。每个数据框都有一个数字列,列出所有测量的胡萝卜(总计:100k 胡萝卜)和黄瓜(总计:50k 黄瓜)的长度。
我希望在同一个图上绘制两个直方图——胡萝卜长度和黄瓜长度。它们重叠,所以我想我还需要一些透明度。我还需要使用相对频率而不是绝对数字,因为每组中的实例数量不同。
这样的东西会很好,但是我如何从我的两个表中创建它?
这是一个使用基础图形和 alpha 混合的更简单的解决方案(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:由于这刚刚获得了赞成票,我想我也可以添加代码生成的视觉效果,因为 alpha 混合非常有用:
您链接到的图像是密度曲线,而不是直方图。
如果您一直在阅读 ggplot,那么您可能唯一缺少的就是将两个数据框组合成一个长数据框。
因此,让我们从您拥有的两组独立数据开始,然后将它们组合起来。
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)
之后,如果您的数据已经是长格式,则无需这样做,您只需要一行即可绘制图表。
ggplot(vegLengths, aes(length, fill = veg)) + geom_density(alpha = 0.2)
现在,如果您确实想要直方图,则以下内容将起作用。请注意,您必须更改默认“stack”参数的位置。如果您并不真正了解数据应该是什么样子,您可能会错过这一点。阿尔法值越高看起来越好。另请注意,我将其制作为密度直方图。删除
y = ..density..
即可轻松恢复计数。
ggplot(vegLengths, aes(length, fill = veg)) +
geom_histogram(alpha = 0.5, aes(y = ..density..), position = 'identity')
另外,我对 Dirk 的问题进行了评论,即所有参数都可以简单地位于
hist
命令中。有人问我如何做到这一点。接下来产生的正是德克的图形。
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)
这是我写的一个函数,使用伪透明度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$counts[i] = 0
}
}
plot(ahist, xlim=xlim, ylim=ylim, col=colors[1])
plot(bhist, xlim=xlim, ylim=ylim, col=colors[2], add=T)
plot(overlap, xlim=xlim, ylim=ylim, col=colors[3], add=T)
}
a=rnorm(1000, 3, 1)
b=rnorm(1000, 6, 1)
hist(a, xlim=c(0,10), col="red")
hist(b, add=T, col=rgb(0, 1, 0, 0.5) )
结果最终看起来像这样:
已经有了漂亮的答案,但我想添加这个。在我看来很好。 (从@Dirk复制随机数)。需要
library(scales)
`
set.seed(42)
hist(rnorm(500,4),xlim=c(0,10),col='skyblue',border=F)
hist(rnorm(500,6),add=T,col=scales::alpha('red',.5),border=F)
结果是...
更新:这个重叠功能可能对某些人也有用。
hist0 <- function(...,col='skyblue',border=T) hist(...,col=col,border=border)
我觉得
hist0
的结果比hist
更漂亮
hist2 <- function(var1, var2,name1='',name2='',
breaks = min(max(length(var1), length(var2)),20),
main0 = "", alpha0 = 0.5,grey=0,border=F,...) {
library(scales)
colh <- c(rgb(0, 1, 0, alpha0), rgb(1, 0, 0, alpha0))
if(grey) colh <- c(alpha(grey(0.1,alpha0)), alpha(grey(0.9,alpha0)))
max0 = max(var1, var2)
min0 = min(var1, var2)
den1_max <- hist(var1, breaks = breaks, plot = F)$density %>% max
den2_max <- hist(var2, breaks = breaks, plot = F)$density %>% max
den_max <- max(den2_max, den1_max)*1.2
var1 %>% hist0(xlim = c(min0 , max0) , breaks = breaks,
freq = F, col = colh[1], ylim = c(0, den_max), main = main0,border=border,...)
var2 %>% hist0(xlim = c(min0 , max0), breaks = breaks,
freq = F, col = colh[2], ylim = c(0, den_max), add = T,border=border,...)
legend(min0,den_max, legend = c(
ifelse(nchar(name1)==0,substitute(var1) %>% deparse,name1),
ifelse(nchar(name2)==0,substitute(var2) %>% deparse,name2),
"Overlap"), fill = c('white','white', colh[1]), bty = "n", cex=1,ncol=3)
legend(min0,den_max, legend = c(
ifelse(nchar(name1)==0,substitute(var1) %>% deparse,name1),
ifelse(nchar(name2)==0,substitute(var2) %>% deparse,name2),
"Overlap"), fill = c(colh, colh[2]), bty = "n", cex=1,ncol=3) }
结果
par(mar=c(3, 4, 3, 2) + 0.1)
set.seed(100)
hist2(rnorm(10000,2),rnorm(10000,3),breaks = 50)
是
以下是如何在“经典”R 图形中执行此操作的示例:
## generate some random data
carrotLengths <- rnorm(1000,15,5)
cucumberLengths <- rnorm(200,20,7)
## calculate the histograms - don't plot yet
histCarrot <- hist(carrotLengths,plot = FALSE)
histCucumber <- hist(cucumberLengths,plot = FALSE)
## calculate the range of the graph
xlim <- range(histCucumber$breaks,histCarrot$breaks)
ylim <- range(0,histCucumber$density,
histCarrot$density)
## plot the first graph
plot(histCarrot,xlim = xlim, ylim = ylim,
col = rgb(1,0,0,0.4),xlab = 'Lengths',
freq = FALSE, ## relative, not absolute frequency
main = 'Distribution of carrots and cucumbers')
## plot the second graph on top of this
opar <- par(new = FALSE)
plot(histCucumber,xlim = xlim, ylim = ylim,
xaxt = 'n', yaxt = 'n', ## don't add axes
col = rgb(0,0,1,0.4), add = TRUE,
freq = FALSE) ## relative, not absolute frequency
## add a legend in the corner
legend('topleft',c('Carrots','Cucumbers'),
fill = rgb(1:0,0,0:1,0.4), bty = 'n',
border = NA)
par(opar)
唯一的问题是,如果直方图中断对齐,看起来会更好,这可能需要手动完成(在传递给
hist
的参数中)。
这是类似于 ggplot2 的版本,我仅在基础 R 中提供。我从 @nullglob 复制了一些。
生成数据
carrots <- rnorm(100000,5,2)
cukes <- rnorm(50000,7,2.5)
您不需要像ggplot2那样将其放入数据框中。这种方法的缺点是你必须写出更多的情节细节。好处是你可以掌控更多的剧情细节。
## calculate the density - don't plot yet
densCarrot <- density(carrots)
densCuke <- density(cukes)
## calculate the range of the graph
xlim <- range(densCuke$x,densCarrot$x)
ylim <- range(0,densCuke$y, densCarrot$y)
#pick the colours
carrotCol <- rgb(1,0,0,0.2)
cukeCol <- rgb(0,0,1,0.2)
## plot the carrots and set up most of the plot parameters
plot(densCarrot, xlim = xlim, ylim = ylim, xlab = 'Lengths',
main = 'Distribution of carrots and cucumbers',
panel.first = grid())
#put our density plots in
polygon(densCarrot, density = -1, col = carrotCol)
polygon(densCuke, density = -1, col = cukeCol)
## add a legend in the corner
legend('topleft',c('Carrots','Cucumbers'),
fill = c(carrotCol, cukeCol), bty = 'n',
border = NA)
@Dirk Eddelbuettel:基本思想非常好,但所示代码可以改进。 [需要很长时间才能解释,因此需要单独回答而不是评论。]
hist()
函数默认绘制绘图,因此您需要添加plot=FALSE
选项。此外,通过 plot(0,0,type="n",...)
调用来建立绘图区域会更清晰,您可以在其中添加轴标签、绘图标题等。最后,我想提一下,还可以使用阴影来区分两个直方图。这是代码:
set.seed(42)
p1 <- hist(rnorm(500,4),plot=FALSE)
p2 <- hist(rnorm(500,6),plot=FALSE)
plot(0,0,type="n",xlim=c(0,10),ylim=c(0,100),xlab="x",ylab="freq",main="Two histograms")
plot(p1,col="green",density=10,angle=135,add=TRUE)
plot(p2,col="blue",density=10,angle=45,add=TRUE)
这是结果(由于 RStudio 有点太宽了 :-) ):
Plotly 的 R API 可能对您有用。下图是这里。
library(plotly)
#add username and key
p <- plotly(username="Username", key="API_KEY")
#generate data
x0 = rnorm(500)
x1 = rnorm(500)+1
#arrange your graph
data0 = list(x=x0,
name = "Carrots",
type='histogramx',
opacity = 0.8)
data1 = list(x=x1,
name = "Cukes",
type='histogramx',
opacity = 0.8)
#specify type as 'overlay'
layout <- list(barmode='overlay',
plot_bgcolor = 'rgba(249,249,251,.85)')
#format response, and use 'browseURL' to open graph tab in your browser.
response = p$plotly(data0, data1, kwargs=list(layout=layout))
url = response$url
filename = response$filename
browseURL(response$url)
完全披露:我在团队中。
有很多很棒的答案,但由于我刚刚编写了一个函数(
'basicPlotteR'包中的
plotMultipleHistograms()
)函数来执行此操作,我想我会添加另一个答案。
此函数的优点是它会自动设置适当的 X 和 Y 轴限制,并定义在所有分布中使用的一组通用 bin。
使用方法如下:
# 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()
函数可以采用任意数量的分布,并且所有通用绘图参数都应适用于它(例如:las
、main
等)。