使用 str_replace_all() 替换为完美匹配

Question

我想替换字符，但它们在一列中重复出现。我可以用完美匹配替换它们吗？

time <-data.frame(c(0,0,1,1,2,2,9,9,10,10,11,11))
names(time)<-c("time")
 time %>%
  dplyr::mutate(year = stringr::str_replace_all(time, pattern = c("0" = "1994",
                                                         "1" = "1995",
                                                         "2" = "1996",
                                                         "9" = "2003",
                                                         "10" = "2004",
                                                         "11" = "2005")))

   time                          year
1     0           1200320035200320034
2     0           1200320035200320034
3     1                    1200320035
4     1                    1200320035
5     2                    1200320036
6     2                    1200320036
7     9                          2003
8     9                          2003
9    10 12003200351200320035200320034
10   10 12003200351200320035200320034
11   11          12003200351200320035
12   11          12003200351200320035

Answer 1

df1 <- transform(time, year = time + 1994)
df1
   time year
1     0 1994
2     0 1994
3     1 1995
4     1 1995
5     2 1996
6     2 1996
7     9 2003
8     9 2003
9    10 2004
10   10 2004
11   11 2005
12   11 2005

Answer 2

替代方法：

library(tidyverse)

time_df <-data.frame(c(0,0,1,1,2,2,9,9,10,10,11,11))
names(time_df)<-c("time")

replacement_df <- tribble(
    ~key, ~time,    
    "0" , "1994",
    "1" , "1995",
    "2" , "1996",
    "9" , "2003",
    "10" , "2004",
    "11" , "2005")

time_df <- mutate(time_df, time = as.character(time))

# Using join --------------------------------------------------------------


#both columns will be type character after join
inner_join(time_df, replacement_df, by = c('time' = 'key')) %>% 
    mutate_all(as.numeric) #set all variables to numeric (optional)
#>    time time.y
#> 1     0   1994
#> 2     0   1994
#> 3     1   1995
#> 4     1   1995
#> 5     2   1996
#> 6     2   1996
#> 7     9   2003
#> 8     9   2003
#> 9    10   2004
#> 10   10   2004
#> 11   11   2005
#> 12   11   2005


# using str_replace with purrr:map---------------------------------------------------

replacement <- c("0" = "1994",
                 "1" = "1995",
                 "2" = "1996",
                 "9" = "2003",
                 "10" = "2004",
                 "11" = "2005")
time_df %>% 
    mutate(time = map_chr(time_df$time, ~str_replace(string = .x, .x, replacement[.x])))
#>    time
#> 1  1994
#> 2  1994
#> 3  1995
#> 4  1995
#> 5  1996
#> 6  1996
#> 7  2003
#> 8  2003
#> 9  2004
#> 10 2004
#> 11 2005
#> 12 2005

^{由 reprex package (v2.0.0)}

创建于 2021-06-21

使用 str_replace_all() 替换为完美匹配

replace with perfect matching using str_replace_all()

r

str-replace