R 按规定间隔切割

在R中是否有某种方式可以在不中断的情况下按规定的间隔进行切割

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例如，如果我想要精确间隔[1,10]中的值；默认情况下，

cut

将此间隔拆分为更小的间隔。

要拆分为预定义的间隔，可以使用

breaks

参数指定一个中断向量

定义一些数据：

x <- sample(0:20, 100, replace=TRUE)
x

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像这样的？从0到1的每0.2的打断

> a <- runif(100)
> cut(a, seq(from = 0, to = 1, by = 0.2))
  [1] (0,0.2]   (0.8,1]   (0.8,1]   (0.6,0.8] (0.6,0.8] (0,0.2]   (0.6,0.8]
  [8] (0.2,0.4] (0.8,1]   (0.4,0.6] (0.8,1]   (0.4,0.6] (0.8,1]   (0.6,0.8]
 [15] (0.8,1]   (0,0.2]   (0.8,1]   (0.8,1]   (0.6,0.8] (0.6,0.8] (0.2,0.4]
 [22] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0,0.2]  
 [29] (0,0.2]   (0.2,0.4] (0,0.2]   (0,0.2]   (0,0.2]   (0,0.2]   (0,0.2]  
 [36] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.8,1]   (0.2,0.4] (0.4,0.6]
 [43] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0.4,0.6]
 [50] (0.6,0.8] (0.6,0.8] (0,0.2]   (0.2,0.4] (0.8,1]   (0.8,1]   (0.8,1]  
 [57] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0,0.2]   (0.8,1]   (0.8,1]   (0.2,0.4]
 [64] (0.8,1]   (0.2,0.4] (0.4,0.6] (0.8,1]   (0,0.2]   (0.4,0.6] (0,0.2]  
 [71] (0.4,0.6] (0.8,1]   (0.6,0.8] (0.4,0.6] (0,0.2]   (0.2,0.4] (0.4,0.6]
 [78] (0,0.2]   (0.2,0.4] (0.8,1]   (0,0.2]   (0.4,0.6] (0.8,1]   (0,0.2]  
 [85] (0,0.2]   (0.2,0.4] (0.2,0.4] (0.4,0.6] (0.8,1]   (0.2,0.4] (0,0.2]  
 [92] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0.8,1]   (0.2,0.4] (0.4,0.6] (0,0.2]  
 [99] (0,0.2]   (0,0.2]  
Levels: (0,0.2] (0.2,0.4] (0.4,0.6] (0.6,0.8] (0.8,1]

>切割（a，顺序（从=0到=1，由=0.2）
[1] (0,0.2]   (0.8,1]   (0.8,1]   (0.6,0.8] (0.6,0.8] (0,0.2]   (0.6,0.8]
[8] (0.2,0.4] (0.8,1]   (0.4,0.6] (0.8,1]   (0.4,0.6] (0.8,1]   (0.6,0.8]
[15] (0.8,1]   (0,0.2]   (0.8,1]   (0.8,1]   (0.6,0.8] (0.6,0.8] (0.2,0.4]
[22] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0,0.2]  
[29] (0,0.2]   (0.2,0.4] (0,0.2]   (0,0.2]   (0,0.2]   (0,0.2]   (0,0.2]  
[36] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.8,1]   (0.2,0.4] (0.4,0.6]
[43] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0.4,0.6]
[50] (0.6,0.8] (0.6,0.8] (0,0.2]   (0.2,0.4] (0.8,1]   (0.8,1]   (0.8,1]  
[57] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0,0.2]   (0.8,1]   (0.8,1]   (0.2,0.4]
[64] (0.8,1]   (0.2,0.4] (0.4,0.6] (0.8,1]   (0,0.2]   (0.4,0.6] (0,0.2]  
[71] (0.4,0.6] (0.8,1]   (0.6,0.8] (0.4,0.6] (0,0.2]   (0.2,0.4] (0.4,0.6]
[78] (0,0.2]   (0.2,0.4] (0.8,1]   (0,0.2]   (0.4,0.6] (0.8,1]   (0,0.2]  
[85] (0,0.2]   (0.2,0.4] (0.2,0.4] (0.4,0.6] (0.8,1]   (0.2,0.4] (0,0.2]  
[92] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0.8,1]   (0.2,0.4] (0.4,0.6] (0,0.2]  
[99] (0,0.2]   (0,0.2]  
级别：（0,0.2]（0.2,0.4）（0.4,0.6）（0.6,0.8）（0.8,1）

我有一个关于“削减”的问题削减（每加仑x$4）给了我水平：（10.4,16.3]（16.3,22.1）（22.1,28）（28,33.9）我们如何提取截止点？这里的截止点是10.4,16.3,22.1,28,33。9@darkage这是一个不同的问题，与原来的问题没有真正的联系。但您的问题在最后一个

？cut

示例中得到了回答。如果这没有帮助，请提出一个新问题。

> a <- runif(100)
> cut(a, seq(from = 0, to = 1, by = 0.2))
  [1] (0,0.2]   (0.8,1]   (0.8,1]   (0.6,0.8] (0.6,0.8] (0,0.2]   (0.6,0.8]
  [8] (0.2,0.4] (0.8,1]   (0.4,0.6] (0.8,1]   (0.4,0.6] (0.8,1]   (0.6,0.8]
 [15] (0.8,1]   (0,0.2]   (0.8,1]   (0.8,1]   (0.6,0.8] (0.6,0.8] (0.2,0.4]
 [22] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0,0.2]  
 [29] (0,0.2]   (0.2,0.4] (0,0.2]   (0,0.2]   (0,0.2]   (0,0.2]   (0,0.2]  
 [36] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.8,1]   (0.2,0.4] (0.4,0.6]
 [43] (0.4,0.6] (0.6,0.8] (0.2,0.4] (0.6,0.8] (0.6,0.8] (0.6,0.8] (0.4,0.6]
 [50] (0.6,0.8] (0.6,0.8] (0,0.2]   (0.2,0.4] (0.8,1]   (0.8,1]   (0.8,1]  
 [57] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0,0.2]   (0.8,1]   (0.8,1]   (0.2,0.4]
 [64] (0.8,1]   (0.2,0.4] (0.4,0.6] (0.8,1]   (0,0.2]   (0.4,0.6] (0,0.2]  
 [71] (0.4,0.6] (0.8,1]   (0.6,0.8] (0.4,0.6] (0,0.2]   (0.2,0.4] (0.4,0.6]
 [78] (0,0.2]   (0.2,0.4] (0.8,1]   (0,0.2]   (0.4,0.6] (0.8,1]   (0,0.2]  
 [85] (0,0.2]   (0.2,0.4] (0.2,0.4] (0.4,0.6] (0.8,1]   (0.2,0.4] (0,0.2]  
 [92] (0.6,0.8] (0.2,0.4] (0.2,0.4] (0.8,1]   (0.2,0.4] (0.4,0.6] (0,0.2]  
 [99] (0,0.2]   (0,0.2]  
Levels: (0,0.2] (0.2,0.4] (0.4,0.6] (0.6,0.8] (0.8,1]