一种dplyr方法是这样使用recode：

library(dplyr)

dfa |> 
  mutate(id = recode(id, !!!setNames(dfc$primary, dfc$colname)))
#>   name id
#> 1 jack  b
#> 2 rose  a

或者第二选项将是使用left_join：

dfa |> 
  left_join(dfc, by = c("id" = "colname")) |> 
  select(name, id = primary)
#>   name id
#> 1 jack  b
#> 2 rose  a

数据

dfa <- data.frame(
              name = c("jack", "rose"),
                id = c(1L, 2L)
)

dfc <- data.frame(
           primary = c("a", "b"),
           colname = c(2L, 1L)
)

我不明白这三个数据集之间的关系是不是这样的：dfa$id等于dfc$colname，dfc$primary等于dfb$id，但如果是这样，可能的答案如下：

# Load the dataframes
dfa <- data.frame(name = c("jack", "rose"),
                  id = c(1, 2))
dfb <- data.frame(id = c("a", "b"),
                  job = c("student", "tutor"))
dfc <- data.frame(primary = c("a", "b"),
                  colname = c(2, 1))

# Obtain the indexes that match between dfa and dfc
index <- match(dfa$id, dfc$colname)

# Set the id from dfa as the id in dfb taking into account the previous indices
dfa$id <- dfb$id[index]

# Print results
dfa

  name id
1 jack  b
2 rose  a