R：如何匹配或过滤具有相同字符串但不同序列的变量？答案

【问题标题】：R: How to match or filter variables with same character strings but different sequence?R：如何匹配或过滤具有相同字符串但不同序列的变量？
【发布时间】：2020-12-14 03:08:44
【问题描述】：

我有一个数据集，其中包含两个由全名（姓名和姓氏）组成的变量。但是，这两个变量的顺序不同：

variable1 由
variable2 由

如何过滤行使得variable1 = variable2？或者我可以修改变量2的顺序以匹配变量1的顺序吗？

我创建了一个小样本来复制数据集（请注意，一些全名包含 3 个或更多单词）：

library(tidyverse)

name_surname <- c("John Smith One", "Jane Smith Two", "John Doe", "Nick Doe", "Chris Froome", "Van den Broeck", "Lance", "Van Dae Le Phillipe")

surname_name <- c("Smith One John", "Smith Two Jane", "Doe John", "Nick Doe", "Froome Chris", "Broeck Van den", "Lance", "Phillipe Van Dae Le")

tibble <- tibble(variable1 = name_surname, variable2 = surname_name)

tibble
#> # A tibble: 8 x 2
#>   variable1           variable2          
#>   <chr>               <chr>              
#> 1 John Smith One      Smith One John     
#> 2 Jane Smith Two      Smith Two Jane     
#> 3 John Doe            Doe John           
#> 4 Nick Doe            Nick Doe           
#> 5 Chris Froome        Froome Chris       
#> 6 Van den Broeck      Broeck Van den     
#> 7 Lance               Lance              
#> 8 Van Dae Le Phillipe Phillipe Van Dae Le

^{由reprex package (v0.3.0) 于 2020 年 8 月 25 日创建}

会话信息

devtools::session_info()
#> ─ Session info ───────────────────────────────────────────────────────────────
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#>  language (EN)                        
#>  collate  en_AU.UTF-8                 
#>  ctype    en_AU.UTF-8                 
#>  tz       Australia/Melbourne         
#>  date     2020-08-25                  
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【问题讨论】：

标签： r string filter character

【解决方案1】：

在空间上拆分变量并根据variable1对variable2进行排序。

tibble$variable3 <- mapply(function(x, y) paste(y[match(x, y)], collapse = " "), 
    strsplit(tibble$variable1, '\\s+'), strsplit(tibble$variable2, '\\s+'))

tibble
# A tibble: 8 x 3
#  variable1           variable2           variable3          
#  <chr>               <chr>               <chr>              
#1 John Smith One      Smith One John      John Smith One     
#2 Jane Smith Two      Smith Two Jane      Jane Smith Two     
#3 John Doe            Doe John            John Doe           
#4 Nick Doe            Nick Doe            Nick Doe           
#5 Chris Froome        Froome Chris        Chris Froome       
#6 Van den Broeck      Broeck Van den      Van den Broeck     
#7 Lance               Lance               Lance              
#8 Van Dae Le Phillipe Phillipe Van Dae Le Van Dae Le Phillipe

创建了一个新变量 (variable3) 用于比较，如果需要，您可以覆盖 tibble 中的 variable2。

【讨论】：

【解决方案2】：

与@Ronak Shah 类似的逻辑，但使用dplyr 和tidyr：

tibble %>%
 rowid_to_column() %>%
 separate_rows(variable1, variable2) %>%
 group_by(rowid) %>%
 mutate(variable2 = variable2[match(variable1, variable2)]) %>%
 summarise(across(starts_with("variable"), paste, collapse = " "))

  rowid variable1           variable2          
  <int> <chr>               <chr>              
1     1 John Smith One      John Smith One     
2     2 Jane Smith Two      Jane Smith Two     
3     3 John Doe            John Doe           
4     4 Nick Doe            Nick Doe           
5     5 Chris Froome        Chris Froome       
6     6 Van den Broeck      Van den Broeck     
7     7 Lance               Lance              
8     8 Van Dae Le Phillipe Van Dae Le Phillipe

【讨论】：