如何根据 TOOL 列中的不同值在 ITEM 上使用多个分隔符？

Question

我有一个这样的数据框

ID <- c("A","A","A","A","A","A","A","A","A")
TOOL <- c("ABC_01","ABC_04","ABC_02",
          "DEF_01","DEF_05","DEF_02",
          "IJK_04","IJK_01","IJK_02") 
ITEM <- c("RFALT.2SHEET.S13","RFTACU_789.L_PM.2_N13","CS20_789VIS.200_L_PM.STD",
          "ACRF55_16_T37\AT_PM3\1 N1","RFALT\2SHEET\S13","RFNF_1_2U\L_PM3\5 N13",
          "RFALT 2SHEET S13","CS20_STD 2DUB_L SP C9","RFNC_DBS_T2000 L_EDGE 1 N13") 

df <- data.frame(ID,TOOL,ITEM,stringsAsFactors = F)

我正在尝试使用基于 TOOLs 列的多个定界符

如果 TOOL LIKE 'ABC'，则提取 ITEM 中的字符串，直到 .（. 是定界符）
如果TOOL LIKE 'DEF'，则提取ITEM中的字符串直到\（\是分隔符）
如果TOOL LIKE 'IJK'，则提取ITEM中的字符串直到</code>（space是分隔符）</li> </ol> <p><strong>期望输出</strong></p> <pre><code> ID TOOL ITEM A ABC_01 RFALT A ABC_04 RFTACU_789 A ABC_02 CS20_789VIS A DEF_01 ACRF55_16_T37 A DEF_05 RFALT A DEF_02 RFNF_1_2U A IJK_04 RFALT A IJK_01 CS20_STD A IJK_02 RFNC_DBS_T2000
我正在尝试使用 dplyr & stringr 包
```
library(dplyr)
library(stringr)
df2 <- df %>%
  filter(grepl("ABC",TOOL)) %>% 
  mutate(ITEM = str_extract(ITEM, "^.*(?=.\)")) %>% 
  filter(grepl("DEF",TOOL)) %>% 
  mutate(ITEM = str_extract(ITEM, "^.*(?=.\.)"))
```
这行不通。有人能给我指出正确的方向吗？

Answer 1

我确信有一种更有效的方法可以使用正则表达式执行此操作，但直接的方法是只使用 strsplit 和 select 第一个拆分

library(dplyr)
df %>% 
  mutate(ITEM2 = ITEM %>% strsplit('\.| |\\') %>% sapply(`[[`, 1))

#   ID   TOOL                        ITEM          ITEM2
# 1  A ABC_01            RFALT.2SHEET.S13          RFALT
# 2  A ABC_04       RFTACU_789.L_PM.2_N13     RFTACU_789
# 3  A ABC_02    CS20_789VIS.200_L_PM.STD    CS20_789VIS
# 4  A DEF_01 ACRF55_16_T37\AT_PM3\1 N1  ACRF55_16_T37
# 5  A DEF_05          RFALT\2SHEET\S13          RFALT
# 6  A DEF_02     RFNF_1_2U\L_PM3\5 N13      RFNF_1_2U
# 7  A IJK_04            RFALT 2SHEET S13          RFALT
# 8  A IJK_01       CS20_STD 2DUB_L SP C9       CS20_STD
# 9  A IJK_02 RFNC_DBS_T2000 L_EDGE 1 N13 RFNC_DBS_T2000

Answer 2

一个选项 str_remove

library(tidyverse)
df %>%
     mutate(ITEM = str_remove(ITEM, "[.\\ ].*"))
#  ID   TOOL           ITEM
#1  A ABC_01          RFALT
#2  A ABC_04     RFTACU_789
#3  A ABC_02    CS20_789VIS
#4  A DEF_01  ACRF55_16_T37
#5  A DEF_05          RFALT
#6  A DEF_02      RFNF_1_2U
#7  A IJK_04          RFALT
#8  A IJK_01       CS20_STD
#9  A IJK_02 RFNC_DBS_T2000

如果模式特定于特定 'TOOL'，一种选择是单独应用 str_remove

map2_df(c("ABC", "DEF", "IJK"), c(".", "\\", " "), ~ 
      df %>% 
         filter(str_detect(TOOL, .x)) %>%
         mutate(ITEM = str_remove(ITEM, paste0("[", .y, "].*"))))
#  ID   TOOL           ITEM
#1  A ABC_01          RFALT
#2  A ABC_04     RFTACU_789
#3  A ABC_02    CS20_789VIS
#4  A DEF_01  ACRF55_16_T37
#5  A DEF_05          RFALT
#6  A DEF_02      RFNF_1_2U
#7  A IJK_04          RFALT
#8  A IJK_01       CS20_STD
#9  A IJK_02 RFNC_DBS_T2000

Answer 3

我们可以使用 case_when 和 str_replace。

library(tidyverse)

df2 <- df %>%
  mutate(ITEM = case_when(
    str_detect(TOOL, "^ABC")     ~str_replace(ITEM, "[\.].*", ""),
    str_detect(TOOL, "^DEF")     ~str_replace(ITEM, "[\\].*", ""),
    str_detect(TOOL, "^IJK")      ~str_replace(ITEM, "[:space:].*", "")
  ))
df2
#   ID   TOOL           ITEM
# 1  A ABC_01          RFALT
# 2  A ABC_04     RFTACU_789
# 3  A ABC_02    CS20_789VIS
# 4  A DEF_01  ACRF55_16_T37
# 5  A DEF_05          RFALT
# 6  A DEF_02      RFNF_1_2U
# 7  A IJK_04          RFALT
# 8  A IJK_01       CS20_STD
# 9  A IJK_02 RFNC_DBS_T2000

如何根据 TOOL 列中的不同值在 ITEM 上使用多个分隔符？

How to use multiple delimiters on ITEM based on different values in TOOL column?

r

stringr

dplyr

data.table

tidyverse