R Studio：匹配两列之间的前n个字符，并从另一列填充值答案

【问题标题】：R Studio: Match first n characters between two columns, and fill in value from another columnR Studio：匹配两列之间的前n个字符，并从另一列填充值
【发布时间】：2021-05-04 00:21:38
【问题描述】：

我有一个如下所示的数据框“city_table”：

+---+---------------------+
|   | city                |
+---+---------------------+
| 1 | Chicago-2234dxsw    |
+---+---------------------+
| 2 | Chicago,IL          |
+---+---------------------+
| 3 | Chicago             |
+---+---------------------+
| 4 | Chicago - 124421xsd |
+---+---------------------+
| 5 | Chicago_2133xx      |
+---+---------------------+
| 6 | Atlanta- 1234xx     |
+---+---------------------+
| 7 | Atlanta, GA         |
+---+---------------------+
| 8 | Atlanta - 123456T   |
+---+---------------------+

我有另一个城市代码查找表“city_lookup”，如下所示：

+---+--------------+-----------+
|   | city_name    | city_code |
+---+--------------+-----------+
| 1 | Chicago, IL  | 001       |
+---+--------------+-----------+
| 2 | Atlanta, GA  | 002       |
+---+--------------+-----------+

如您所见，“city”中的城市名称混乱且格式不同，而“city_code”中的城市名称遵循统一格式（city,STATE）。

我希望决赛桌通过匹配city_table$city 与city_lookup$city_name 之间的前n 个字符（让我们看看，n=7），返回城市代码正确的，像这样：

+---+---------------------+-----------+
|   | city_name           | city_code |
+---+---------------------+-----------+
| 1 | Chicago-2234dxsw    | 001       |
+---+---------------------+-----------+
| 2 | Chicago,IL          | 001       |
+---+---------------------+-----------+
| 3 | Chicago             | 001       |
+---+---------------------+-----------+
| 4 | Chicago - 124421xsd | 001       |
+---+---------------------+-----------+
| 5 | Chicago_2133xx      | 001       |
+---+---------------------+-----------+
| 6 | Atlanta- 1234xx     | 002       |
+---+---------------------+-----------+
| 7 | Atlanta, GA         | 002       |
+---+---------------------+-----------+
| 8 | Atlanta - 123456T   | 002       |
+---+---------------------+-----------+

我在 R 中执行此操作，最好使用 tidyverse/dplyr。非常感谢您的帮助！

【问题讨论】：

标签： r dplyr tidyverse

【解决方案1】：

更好的是，只要城市全名后面的字符都是非字母的，就可以这样匹配整个城市名：

city_table <- tibble(city = c("Chicago-2234dxsw", "Chicago,IL", "Atlanta - 123456T"))
city_lookup <- tibble(city_name = c("Chicago, IL", "Atlanta, GA"),
                      city_code = c("001", "002"))


city_table %>%
  mutate(city_clean  = gsub("^([a-zA-Z]*).*", "\\1", city)) %>%
  left_join(city_lookup %>%
              mutate(city_clean  = gsub("^([a-zA-Z]*).*", "\\1", city_name, perl = T)),
            by = "city_clean") %>%
  select(-city_clean, -city_name)


  city              city_code
  <chr>             <chr>    
1 Chicago-2234dxsw  001      
2 Chicago,IL        001      
3 Atlanta - 123456T 002

【讨论】：

【解决方案2】：

我们可以使用 substring 创建列（正如问题中的 OP 所问），然后执行 regex_left_join

library(dplyr)
library(fuzzyjoin)
city_table %>%
   mutate(city_sub = substring(city, 1, 7)) %>%
   regex_left_join(city_lookup %>%
                     mutate(city_sub = substring(city_name, 1, 7)), 
             by = 'city_sub')  %>%
   select(city_name = city, city_code)

-输出

#             city_name city_code
#1    Chicago-2234dxsw       001
#2          Chicago,IL       001
#3             Chicago       001
#4 Chicago - 124421xsd       001
#5      Chicago_2133xx       001
#6     Atlanta- 1234xx       002
#7         Atlanta, GA       002
#8   Atlanta - 123456T       002

数据

city_table <- structure(list(city = c("Chicago-2234dxsw", "Chicago,IL", "Chicago", 
"Chicago - 124421xsd", "Chicago_2133xx", "Atlanta- 1234xx", "Atlanta, GA", 
"Atlanta - 123456T")), class = "data.frame", row.names = c(NA, 
-8L))

city_lookup <- structure(list(city_name = c("Chicago, IL", "Atlanta, GA"), 
city_code = c("001", 
"002")), class = "data.frame", row.names = c(NA, -2L))

【讨论】：