根据数值变量的水平对新因子进行编码

Question

我正在尝试根据另一列的数值创建一个因子列。这是我的数据的一个子集：

> dput(sample)
structure(list(ID = c(1683L, 1684L, 1684L, 1684L, 1684L, 1685L, 
1685L, 1685L, 1685L, 1686L, 1686L, 1686L, 1686L, 30759L, 30759L, 
30759L, 30759L, 30760L, 30760L, 30760L, 30760L), Month = structure(c(2L, 
2L, 3L, 1L, 2L, 2L, 3L, 1L, 2L, 2L, 3L, 1L, 2L, 2L, 3L, 1L, 2L, 
2L, 3L, 1L, 2L), .Label = c("Jun", "Jul", "Aug"), class = "factor"), 
    Year = c(2018, 2017, 2017, 2018, 2018, 2017, 2017, 2018, 
    2018, 2017, 2017, 2018, 2018, 2017, 2017, 2018, 2018, 2017, 
    2017, 2018, 2018), Homerange = c(NA, 27.2850594918174, NA, 
    NA, NA, NA, 30.52684873837, NA, NA, NA, 30.7069481409563, 
    10.625864752589, 29.2661529202662, 32.3278427642325, NA, 
    NA, NA, NA, 33.8586876862157, NA, NA)), out.attrs = list(
    dim = c(58L, 4L, 2L), dimnames = list(Var1 = c("Var1= 1657", 
    "Var1= 1658", "Var1= 1659", "Var1= 1660", "Var1= 1661", "Var1= 1662", 
    "Var1= 1663", "Var1= 1664", "Var1= 1666", "Var1= 1667", "Var1= 1668", 
    "Var1= 1669", "Var1= 1670", "Var1= 1671", "Var1= 1672", "Var1= 1673", 
    "Var1= 1674", "Var1= 1675", "Var1= 1676", "Var1= 1678", "Var1= 1679", 
    "Var1= 1680", "Var1= 1681", "Var1= 1682", "Var1= 1683", "Var1= 1684", 
    "Var1= 1685", "Var1= 1686", "Var1=30759", "Var1=30760", "Var1=30761", 
    "Var1=30762", "Var1=30763", "Var1=30764", "Var1=30765", "Var1=30766", 
    "Var1=30767", "Var1=30768", "Var1=30769", "Var1=30770", "Var1=30771", 
    "Var1=30772", "Var1=30773", "Var1=30774", "Var1=30775", "Var1=30776", 
    "Var1=30777", "Var1=30778", "Var1=30779", "Var1=30780", "Var1=30781", 
    "Var1=30782", "Var1=30783", "Var1=30784", "Var1=30785", "Var1=30786", 
    "Var1=30787", "Var1=30788"), Var2 = c("Var2=Jun", "Var2=Jul", 
    "Var2=Aug", "Var2=Sep"), Var3 = c("Var3=2017", "Var3=2018"
    ))), row.names = c(315L, 84L, 142L, 258L, 316L, 85L, 143L, 
259L, 317L, 86L, 144L, 260L, 318L, 87L, 145L, 261L, 319L, 88L, 
146L, 262L, 320L), class = "data.frame")

数字列 "ID" 的值介于 1659-1685 和 30759-30788 之间。我想做的是创建一个因子列 "Type"，它有 2 个级别 "V13"，对应于 ID 1659-1685，"V16" 对应于 ID 30759-30788。我知道我以前做过，但出于某种原因我不记得是怎么做到的。感谢您的帮助！

Answer 1

假设 ID 1686 是故意不在您的范围内的，您可以试试这个：

library(dplyr)
library(forcats)
df %>% 
  mutate(type = case_when(between(ID, 1659, 1685) ~ "V13",
                          between(ID, 30759, 30788) ~ "V16")) %>%
  mutate(type = as_factor(type))

# A tibble: 21 x 5
      ID Month  Year Homerange type 
   <int> <fct> <dbl>     <dbl> <fct>
 1  1683 Jul    2018      NA   V13  
 2  1684 Jul    2017      27.3 V13  
 3  1684 Aug    2017      NA   V13  
 4  1684 Jun    2018      NA   V13  
 5  1684 Jul    2018      NA   V13  
 6  1685 Jul    2017      NA   V13  
 7  1685 Aug    2017      30.5 V13  
 8  1685 Jun    2018      NA   V13  
 9  1685 Jul    2018      NA   V13  
10  1686 Jul    2017      NA   NA   
11  1686 Aug    2017      30.7 NA   
12  1686 Jun    2018      10.6 NA   
13  1686 Jul    2018      29.3 NA   
14 30759 Jul    2017      32.3 V16  
15 30759 Aug    2017      NA   V16  
16 30759 Jun    2018      NA   V16  
17 30759 Jul    2018      NA   V16  
18 30760 Jul    2017      NA   V16  
19 30760 Aug    2017      33.9 V16  
20 30760 Jun    2018      NA   V16  
21 30760 Jul    2018      NA   V16

Answer 2

直接基础 R 解决方案将应用 ifelse。

sample <- transform(sample,
                    Type=factor(ifelse(ID %in% 1659:1685, "V13", 
                                       ifelse(ID %in% 30759:30788, "V16",
                                              NA))))

或者使用 cut 稍微更有效（感谢 @camille）：

transform(sample, Type2=cut(sample$ID, c(1659, 1685, 1686, 30788), include.lowest=TRUE,
                            labels=c("V13", NA, "V16")))

或 data.table::inrange

library(data.table)
sample <- transform(sample,
                    Type=factor(ifelse(ID %inrange% c(1659, 1685), "V13", 
                                       ifelse(ID %inrange% c(30759, 30788), "V16",
                                              NA))))

str(sample)
# 'data.frame': 21 obs. of  5 variables:
# $ ID       : int  1683 1684 1684 1684 1684 1685 1685 1685 1685 1686 ...
# $ Month    : Factor w/ 3 levels "Jun","Jul","Aug": 2 2 3 1 2 2 3 1 2 2 ...
# $ Year     : num  2018 2017 2017 2018 2018 ...
# $ Homerange: num  NA 27.3 NA NA NA ...
# $ Type     : Factor w/ 2 levels "V13","V16": 1 1 1 1 1 1 1 1 1 NA ...

根据数值变量的水平对新因子进行编码

Coding a new factor based on levels of a numeric variable

r

factors