Question

我正在尝试根据另一列中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“类型”，其具有2个级别“ V13”，分别对应于ID 1659-1685，而“ V16”则对应于ID 30759-30788。我知道我以前做过，但是由于某种原因，我不记得怎么做。感谢您的帮助！

Answer 1

假设您的范围内没有考虑使用ID 1686是故意的，您可以尝试以下方法：

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

直接基于R的解决方案是应用cut。

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

或更有效的使用data.table::inrange（对 @camille 的赠送金额）：

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

或带有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 ...

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                                     BranchB--CommitB1--CommitB2

Develop
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        Branch2--CommitA1--CommitA2--CommitB1--CommitB2

根据数字变量的级别编码新因子

2 个答案: