r - R中的切割和表格功能

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：

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

如果添加 right=TRUE，则区间为
[30,40) [40,50) [50,60)

我怎样才能得到诸如 [30,40] (40,50] (50,60]或'[30,40）[40,50）[50,60]'之类的间隔？

问题2：

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

我知道区间 (30,40] 中有 23 个数字，我可以用 . 得到它们 x[x<=40 & x>30]。
有没有更好的方法来得到结果？

Answer 1

对于问题 2，只需使用 cut 的输出

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

Answer 2

对于您的问题＃2，这取决于您所说的“更好”是什么意思。

这是一种选择：

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

或者，cut您可以使用findInterval：

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

你也可以看看subset函数。

如果这些不符合您对“更好”的定义，请告诉我们更多关于您想要什么的信息。