Question

x<-sample(30:60,50,TRUE)
y<-cut(x,breaks=c(30,40,50,60))
y
[1] (30,40] (30,40] (50,60] (40,50] (40,50] (40,50] (40,50] (30,40] (30,40]
[10] (50,60] (30,40] (50,60] (30,40] (30,40] (50,60] (50,60] (50,60] (30,40]
[19] (50,60] (30,40] (40,50] (40,50] (30,40] (30,40] (30,40] (40,50] (30,40]
[28] (50,60] (40,50] (40,50] (30,40] (50,60] (40,50] (50,60] (50,60] (30,40]
[37] (50,60] (50,60] (30,40] (50,60] (30,40] (30,40] <NA> (40,50] (30,40]
[46] (40,50] (30,40] (30,40] (30,40] (30,40]
Levels: (30,40] (40,50] (50,60]
table(y)
y
(30,40] (40,50] (50,60]
23 12 14
table(y)[1]
(30,40]
23

问题1：

如果添加right = FALSE，则间隔为
（30,40）（40,50）（50,60）

如果添加right = TRUE，则间隔为
[30,40] [40,50] [50,60]

我怎样才能得到[30,40] (40,50] (50,60]或'[30,40] [40,50] [50,60]的间隔？

问题2：

table(y)[1]  
(30,40]
23

我知道间隔期间有30个数字（30,40），我可以全部用它们 x[x<=40 & x>30]。
有没有更好的方法来获得结果？

Answer 1

对于问题2，只需使用剪切的输出

x[y == "(30,40]"]

Answer 2

对于你的问题＃2，它取决于你对“更好”的意思。

这是一个选项：

library(TeachingDemos)
x[ 30 %<% x %<=% 40 ]

或者，您可以使用cut：

，而不是使用findInterval

y <- findInterval(x, c(30,40,50,60))
x[ y==1 ]

您还可以查看subset功能。

如果这些与“更好”的定义不符，请告诉我们您想要的更多信息。

R中的剪切和表格功能

2 个答案: