【问题标题】：Separate rows on string & preserving original在字符串上分隔行并保留原始行
【发布时间】：2023-03-21 12:28:02
【问题描述】：

我正在做一个项目，我想要特定字符串的所有排列。我正在使用tidyr::separate_rows 在特定字符串上拆分和复制，但我想保留原始行。

require(dplyr)
require(tidyr)
temp <- tibble(raw_name = c("happy bank dba american bank and trust", " sohappy bank dba american bank"), clean_name = c("american bank and trust", "american bank"))

这就是我现在正在做的事情：

final <- temp %>%
    separate_rows(raw_name, sep = "dba")

这会丢失我原来的行。我查看了文档，但没有找到 .keep_all = TRUE 的版本。这是上面separate_rows的结果：

  raw_name                               clean_name             
  <chr>                                  <chr>                  
1 happy bank dba american bank and trust american bank and trust
2 " sohappy bank dba american bank"      american bank

我当前的解决方案是创建一个带有违规观察的新 df，执行 separate_rows 并 rbind 原始行。这是我想要的结果：

raw_name                               clean_name             
  <chr>                                  <chr>                  
1 happy bank dba american bank and trust american bank and trust
2 " sohappy bank dba american bank"      american bank          
3 "happy bank "                          american bank and trust
4 " american bank and trust"             american bank and trust
5 " sohappy bank "                       american bank          
6 " american bank"                       american bank

谢谢大家！

【问题讨论】：

标签： r regex tidyr data-cleaning

【解决方案1】：

行分离后，我们可以与原始数据集进行绑定

library(dplyr)
library(tidyr)
temp %>% 
  separate_rows(raw_name, sep="\\s*dba\\s*") %>%
  bind_rows(temp, .)
# A tibble: 6 x 2
#  raw_name                                 clean_name             
#* <chr>                                    <chr>                  
#1 "happy bank dba american bank and trust" american bank and trust
#2 " sohappy bank dba american bank"        american bank          
#3 "happy bank"                             american bank and trust
#4 "american bank and trust"                american bank and trust
#5 " sohappy bank"                          american bank          
#6 "american bank"                          american bank

【讨论】：

感谢您的成功。您能否为 dba 周围的\\s* \\s* 提供更多背景信息？这只是捕获额外的空间还是在做其他事情？
@Francisco * 意味着零个或多个空格 (\\s) 即。捕捉之前或之后的空间

【解决方案2】：

基于@akrun 的回答，我们可以有另一种方法

temp <- data.frame(raw_name = c("happy bank dba american bank and trust", 
                            " sohappy bank dba american bank"), 
               clean_name = c("american bank and trust", "american bank"), stringsAsFactors = FALSE)

temp2 <- temp %>% 
  mutate(raw_name = strsplit(raw_name, "\\s*dba\\s*")) %>% 
  unnest(raw_name) %>% 
  bind_rows(temp)

输出

  raw_name                                 clean_name             
  <chr>                                    <chr>                  
1 "happy bank"                             american bank and trust
2 "american bank and trust"                american bank and trust
3 " sohappy bank"                          american bank          
4 "american bank"                          american bank          
5 "happy bank dba american bank and trust" american bank and trust
6 " sohappy bank dba american bank"        american bank

更新：如果你想去掉字符串开头的空格，你可以试试：

Library(stringr)
temp2$raw_name <- str_squish(temp2$raw_name)
output:
  raw_name                               clean_name             
  <chr>                                  <chr>                  
1 happy bank                             american bank and trust
2 american bank and trust                american bank and trust
3 sohappy bank                           american bank          
4 american bank                          american bank          
5 happy bank dba american bank and trust american bank and trust
6 sohappy bank dba american bank         american bank

【讨论】：