Question

我已经提供的数据已经包含了类别的总数。我正在尝试为每个com和年龄制作房屋数量的直方图，其中Houses是该类别的总数。

Com<-c( "Newport",  "Newport",  "Newport",  "Newport",  "Newport",  "Newport",  "Topeka",   "Topeka",   "Topeka",   "Topeka",   "Topeka",   "Topeka",   "Missoula", "Missoula", "Missoula", "Missoula", "Missoula", "Missoula"  )
Age<-c( "1970s",    "1960s",    "1950s",    "1940s",    "1940_earlier", "1990s",    "1970s",    "1960s",    "1950s",    "1940s",    "1940_earlier", "1990s",    "1970s",    "1960s",    "1950s",    "1940s",    "1940_earlier", "1990s" )
Houses<-c(  11, 6,  3,  0,  0,  21, 44, 0,  3,  3,  25, 20, 0,  51, 236,    192,    312,    299 )
df=data.frame(Com,Age,Houses)

所以df是数据

histogram( ~ Age | Com, data=df)

我也试过

install.packages("ggplot2")
library(ggplot2) 
g <- ggplot(df$counts, aes(df$Age))
g + geom_bar()

另外

barplot(prop.table(table(df$Age)))

最后

p <- ggplot(data = df, aes(x=Age)) 
p <- p + geom_histogram(aes(weights=Houses, fill=Com))
p <- p + scale_fill_brewer(palette="Set3")
p <- p + facet_wrap( ~ Com, ncol=1)
p

这是我的R版本信息：

R.Version()

$platform
[1] "x86_64-w64-mingw32"
$arch
[1] "x86_64"
$os
[1] "mingw32"
$system
[1] "x86_64, mingw32"
$status
[1] ""
$major
[1] "3"
$minor
[1] "3.0"
$year
[1] "2016"
$month
[1] "05"
$day
[1] "03"
$`svn rev`
[1] "70573"
$language
[1] "R"
$version.string
[1] "R version 3.3.0 (2016-05-03)"
$nickname
[1] "Supposedly Educational"

Answer 1

在geom_bar()中使用已经计算过的总计ggplot2时（而不是需要对案例进行计数或求和），您需要指定stat="identity"。 <怎么样

g0 <- ggplot(df,aes(Age,Houses))+
   geom_bar(stat="identity")+
   facet_wrap(~Com)
print(g0)

？或

ggplot(df,aes(Age,Houses))+
   geom_bar(stat="identity")+
   coord_flip()+
   facet_wrap(~Com,ncol=1)

Answer 2

格子函数是条形图：

library(lattice)
barchart( Houses ~ Age ,group=Com,  data=df)

基本条形图解决方案要求数据位于矩阵，表格或xtabs对象中。通过defatul你得到一个堆栈条形图，但是如果你和上面的格子代码一样，你可以添加旁边的参数：

barplot(xtabs(Houses~Com+Age, data=df), beside=TRUE)

具有计数的类别的直方图

2 个答案: