合并具有多個分隔符的列

如何合并具有可變數量分隔符的列，以便獲得類似輸出的內容（假設所有內容都是字符）？

 dt1
   letter
1       a
2     b c
3       c
4 d e f g
5   a g e

 dt2
  letter number
1      a      1
2      b      2
3      c      3
4      d      4
5      e      5
6      f      6
7      g      7

> output
   letter  number
1       a       1
2     b c     2 3
3       c       3
4 d e f g 4 5 6 7
5   a g e   1 7 5

dt1<-data.frame(letter=c("a","b c","c","d e f g","a g e"))

dt2<-data.frame(letter=c("a","b","c","d","e","f","g"),number=c("1","2","3","4","5","6","7"))

output<-data.frame(letter=c("a","b c","c","d e f g","a g e"), number=c("1","2 3","3","4 5 6 7","1 7 5"))

uj5u.com熱心網友回復：

一個基本的 R 解決方案可以是，

dt1$res <- sapply(strsplit(dt1$letter, ' ', fixed = TRUE), function(i)paste(dt2$number[dt2$letter %in% i], collapse = ' '))

#   letter     res
#1       a       1
#2     b c     2 3
#3       c       3
#4 d e f g 4 5 6 7
#5   a g e   1 5 7

uj5u.com熱心網友回復：

無需拆分任何資料，因為您只想替換特定數字的特定字母。

dt1<-data.frame(letter=c("a","b c","c","d e f g","a g e"), stringsAsFactors = F)
dt2<-data.frame(letter=c("a","b","c","d","e","f","g"),number=c("1","2","3","4","5","6","7"), stringsAsFactors = F)

library(stringi)

dt1 %>% mutate(number = stri_replace_all_regex(letter, dt2$letter, dt2$number, vectorize_all = F))

   letter  number
1       a       1
2     b c     2 3
3       c       3
4 d e f g 4 5 6 7
5   a g e   1 7 5

另一個解決方案可能更短

dt1 <- data.frame(letter=c("a","b c","c","d e f g","a g e"), stringsAsFactors = F)

v <- c("1","2","3","4","5","6","7")
names(v) <- c("a","b","c","d","e","f","g")

dt1 %>% mutate(number = str_replace_all(letter, v))

uj5u.com熱心網友回復：

library(tidyverse)
dt1 <- data.frame(letter = c("a", "b c", "c", "d e f g", "a g e"))
dt2 <- data.frame(letter = c("a", "b", "c", "d", "e", "f", "g"),
  number = c("1", "2", "3", "4", "5", "6", "7"))

dt1 %>%
  as_tibble() %>%
  mutate(
    number = letter %>% map_chr(~ .x %>%
      str_split("[ ]") %>%
      simplify() %>%
      map_chr(~ deframe(dt2)[.x]) %>%
      paste0(collapse = " ")
    )
  )
#> # A tibble: 5 x 2
#>   letter  number
#>   <chr>   <chr>  
#> 1 a       1      
#> 2 b c     2 3    
#> 3 c       3      
#> 4 d e f g 4 5 6 7
#> 5 a g e   1 7 5

^{由reprex 包(v2.0.1)于 2021 年 12 月 14 日創建}

uj5u.com熱心網友回復：

使用tidyverse.

library(tidyverse)

output <- dt1 %>%
  mutate(ID = 1:n()) %>%
  separate_rows(letter, sep = "\\ ") %>%
  left_join(dt2, by = "letter") %>%
  group_by(ID) %>%
  summarize(across(.fns = ~paste0(., collapse = " "))) %>%
  ungroup() %>%
  select(-ID)
output
# # A tibble: 5 x 2
#   letter  number 
#   <chr>   <chr>  
# 1 a       1      
# 2 b c     2 3    
# 3 c       3      
# 4 d e f g 4 5 6 7
# 5 a g e   1 7 5 

uj5u.com熱心網友回復：

dt1<-data.frame(letter=c("a","b c","c","d e f g","a g e"))

dt2<-data.frame(letter=c("a","b","c","d","e","f","g"),number=c("1","2","3","4","5","6","7"))

library(tidyverse)
dt1 %>% 
  rowwise() %>% 
  mutate(tmp = str_split(letter, pattern = "\\ ")) %>% 
  ungroup() %>% 
  mutate(number = map_chr(tmp, ~paste0(match(.x, dt2$letter), collapse = " "))) %>% 
  select(-tmp)
#> # A tibble: 5 x 2
#>   letter  number 
#>   <chr>   <chr>  
#> 1 a       1      
#> 2 b c     2 3    
#> 3 c       3      
#> 4 d e f g 4 5 6 7
#> 5 a g e   1 7 5

^{由reprex 包(v2.0.1)于 2021 年 12 月 14 日創建}

標籤：r dplyr 代替 合并 分裂

上一篇：R中方括號的作用

下一篇：如何使用sf更改共享邊框的顏色？