Question

我想使用cut()中定义的中断来剪切数据：

x = c(-10:10)
cut(x, c(-2,4,6,7))
[1] <NA>   <NA>   <NA>   <NA>   <NA>   <NA>   <NA>   <NA>   <NA>   (-2,4] (-2,4] (-2,4] (-2,4] (-2,4] (-2,4] (4,6]  (4,6]  (6,7]  <NA>   <NA>  
[21] <NA>  
Levels: (-2,4] (4,6] (6,7]

但是，我也想获得级别(minimum:-2]和(7:maximum]。在汽车包装的功能recode()中，可以使用“lo：”。是否有类似的东西可用于切割？

9

6

2

Answer 1

x <- -10:10
cut(x, c(-Inf, -2, 4, 6, 7, +Inf))
# Levels: (-Inf,-2] (-2,4] (4,6] (6,7] (7, Inf]

Answer 2

findInterval是答案。

i <- findInterval(x, c(-2,4,6,7))
cbind(x, i)
        x i
 [1,] -10 0
 [2,]  -9 0
 [3,]  -8 0
 [4,]  -7 0
 [5,]  -6 0
 [6,]  -5 0
 [7,]  -4 0
 [8,]  -3 0
 [9,]  -2 1
[10,]  -1 1
[11,]   0 1
[12,]   1 1
[13,]   2 1
[14,]   3 1
[15,]   4 2
[16,]   5 2
[17,]   6 3
[18,]   7 4
[19,]   8 4
[20,]   9 4
[21,]  10 4

Answer 3

您可以使用min()和max()来评估间隔范围（如 Gavin 提到的），并设置include.lowest = TRUE以确保最小值（这里：-10）是间隔的一部分。

输入:

x = c(-10:10)
cut(x, c(min(x),-2,4,6,7,max(x)), include.lowest = TRUE)

输出：

 [1] [-10,-2] [-10,-2] [-10,-2] [-10,-2] [-10,-2] [-10,-2] [-10,-2] [-10,-2] [-10,-2] (-2,4]  
[11] (-2,4]   (-2,4]   (-2,4]   (-2,4]   (-2,4]   (4,6]    (4,6]    (6,7]    (7,10]   (7,10]  
[21] (7,10]  
Levels: [-10,-2] (-2,4] (4,6] (6,7] (7,10]

Answer 4

我遇到了麻烦填充与Inf＆ amp;-Inf之前（虽然正是为什么在这个小时逃脱我），所以一个更安全的解决方案可能是用最小和最大值垫适当扩展：

x <- c(-10:10)
cut(x, c(min(x) -1 , -2, 4, 6, 7, max(x) + 1))
R> x <- c(-10:10)
R> cut(x, c(min(x) -1 , -2, 4, 6, 7, max(x) + 1))
 [1] (-11,-2] (-11,-2] (-11,-2] (-11,-2] (-11,-2] (-11,-2] (-11,-2] (-11,-2]
 [9] (-11,-2] (-2,4]   (-2,4]   (-2,4]   (-2,4]   (-2,4]   (-2,4]   (4,6]   
[17] (4,6]    (6,7]    (7,11]   (7,11]   (7,11]  
Levels: (-11,-2] (-2,4] (4,6] (6,7] (7,11]

在大多数情况下，斯文的答案 / 解决方案就足够了。