R将多列中的名称替换为ID'；它来自另一个物体

我有一组患者数据

df

我正试图在R中取消识别

structure(list(name = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("Andrew", 
                                                                      "Jim", "Kurt", "Lester", "Mickey", "Taylor"), class = "factor"), 
               heart_rate = c(78L, 82L, 67L, 105L, 85L, 94L), age = c(35L, 
                                                                      23L, 43L, 52L, 33L, 45L), partner = structure(c(5L, 2L, 6L, 
                                                                                                                      1L, 3L, 4L), .Label = c("Andrew", "Jim ", "Kurt ", "Lester ", 
                                                                                                                                              "Mickey ", "Taylor "), class = "factor")), class = "data.frame", row.names = c(NA, 
                                                                                                                                                                                                                             -6L))

我想根据名为

key

的对象的

id

列替换

name

和

partner

列的名称

structure(list(name = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("Andrew", 
"Jim", "Kurt", "Lester", "Mickey", "Taylor"), class = "factor"), 
    id = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("A3", 
    "J9", "K5", "L4", "M4", "T7"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L))

我可以使用此代码取消识别

名称

列

df[["name"]] <- key[ match(df[['name']], key[['name']] ) , 'id']

df[["partner"]] <- key[ match(df[['partner']], key[['name']] ) , 'id']

我的数据框看起来像这样

structure(list(name = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("A3", 
"J9", "K5", "L4", "M4", "T7"), class = "factor"), heart_rate = c(78L, 
82L, 67L, 105L, 85L, 94L), age = c(35L, 23L, 43L, 52L, 33L, 45L
), partner = structure(c(NA, NA, NA, 1L, NA, NA), .Label = c("A3", 
"J9", "K5", "L4", "M4", "T7"), class = "factor")), row.names = c(NA, 
-6L), class = "data.frame")

---

有人有什么建议吗？对于那些可以在一行中应用于数据集中所有列的方法，以及对代码的解释，我们非常感谢

问题是在

df

中的

partner

列中，大多数单词后面都有空格：

.Label = c("Andrew", "Jim ", "Kurt ", "Lester ", "Mickey ", "Taylor ")

这意味着

match（）

将找不到完全匹配的名称，除了名称“Andrew”，它将正确返回该索引

解决这个问题的方法是使用

df$partner = trimws(df$partner)

那么您的代码就可以正常工作了：

> df[["partner"]] <- key[ match(df[['partner']], key[['name']] ) , 'id']
> df
  name heart_rate age partner
1   J9         78  35      M4
2   M4         82  23      J9
3   A3         67  43      T7
4   T7        105  52      A3
5   L4         85  33      K5
6   K5         94  45      L4

>df[[“合作伙伴”]]df
姓名心率年龄伙伴
1 J9 78 35 M4
2 M4 82 23 J9
3 A3 67 43 T7
4 T7 105 52 A3
5 L4 85 33 K5
6K59445L4

问题在于，在

df

中的

合作伙伴
列中，大多数单词后面都有一个空格：

.Label = c("Andrew", "Jim ", "Kurt ", "Lester ", "Mickey ", "Taylor ")

这意味着
match（）
将找不到完全匹配的名称，除了名称“Andrew”，它将正确返回该索引

解决这个问题的方法是使用

df$partner = trimws(df$partner)

那么您的代码就可以正常工作了：

> df[["partner"]] <- key[ match(df[['partner']], key[['name']] ) , 'id']
> df
  name heart_rate age partner
1   J9         78  35      M4
2   M4         82  23      J9
3   A3         67  43      T7
4   T7        105  52      A3
5   L4         85  33      K5
6   K5         94  45      L4

>df[[“合作伙伴”]]df
姓名心率年龄伙伴
1 J9 78 35 M4
2 M4 82 23 J9
3 A3 67 43 T7
4 T7 105 52 A3
5 L4 85 33 K5
6K59445L4