通过R中的部分字符串匹配合并

Question

我有一个 df 如下

+-------+---------+-------+
| Brand |  WORD   | Count |
+-------+---------+-------+
| ABC   | cell    |     1 |
| DEF   | dock    |     2 |
| XYZ   | surface |     3 |
| LMN   | pro     |     4 |
| ABC   | mobile  |     5 |
| DEF   | game    |     6 |
| XYZ   | mouse   |     7 |
+-------+---------+-------+

还有一个：

+-------+-----------------+--------+
| Brand |      Name       | profit |
+-------+-----------------+--------+
| ABC   | cell game       |     10 |
| ABC   | cellular mobile |     20 |
| DEF   | docking station |     30 |
| XYZ   | surface mouse   |     40 |
| XYZ   | mouse device    |     50 |
| LMN   | pro device      |     60 |
+-------+-----------------+--------+

我想通过部分字符串匹配（逐字逐句，意味着单元格仅与单元格而不是蜂窝单元匹配）单词和名称并按品牌分组来合并它们，因此结果表如下：

+-------+---------------+-----------------+-------+--------+
| Brand |     WORD      |      Name       | Count | profit |
+-------+---------------+-----------------+-------+--------+
| ABC   | cell          | cell game       |     1 |     10 |
| ABC   | mobile        | cellular mobile |     5 |     20 |
| XYZ   | surface mouse | surface mouse   |     3 |     40 |
| XYZ   | mouse         | mouse device    |     7 |     50 |
| XYZ   | mouse         | mouse device    |     7 |     50 |
| LMN   | pro           | pro device      |     4 |     60 |
+-------+---------------+-----------------+-------+--------+

我尝试使用这里的解决方案 R partial string matching and return value (in R)

但它甚至匹配字符串的一部分，例如单元格将与蜂窝匹配 想知道是否有办法让字符串完全匹配并以所需的形式获得结果

Answer 1

我们在这里假设您要将Brand 列和WORD 列与Name 列匹配，并且输出将按profit 排序。问题中显示的输出有一个重复的行，我们认为这是一个错误。输入 d1 和 d2 在末尾的注释中重复显示。

我们在WORD 和Name 的两边添加一个空格，以确保只使用单词匹配。 like 模式中使用的 % 是一个通配符，可以匹配任何 0 个或多个字符的字符串。

library(sqldf)

sqldf("select d1.Brand, d2.Name, d1.WORD, d1.Count, d2.profit
  from d1
  join d2 on d1.Brand = d2.Brand and 
             ' ' || d2.Name || ' ' like '% ' || d1.WORD || ' %'
  order by d2.profit")

给予：

  Brand            Name    WORD Count profit
1   ABC       cell game    cell     1     10
2   ABC cellular mobile  mobile     5     20
3   XYZ   surface mouse surface     3     40
4   XYZ   surface mouse   mouse     7     40
5   XYZ    mouse device   mouse     7     50
6   LMN      pro device     pro     4     60

注意

可重现形式的输入。

d1 <-
structure(list(Brand = c("ABC", "DEF", "XYZ", "LMN", "ABC", "DEF", 
"XYZ"), WORD = c("cell", "dock", "surface", "pro", "mobile", 
"game", "mouse"), Count = c(1, 2, 3, 4, 5, 6, 7)), class = "data.frame", row.names = c(NA, 
-7L))

d2 <-
structure(list(Brand = c("ABC", "ABC", "DEF", "XYZ", "XYZ", "LMN"
), Name = c("cell game", "cellular mobile", "docking station", 
"surface mouse", "mouse device", "pro device"), profit = c(10, 
20, 30, 40, 50, 60)), class = "data.frame", row.names = c(NA, 
-6L))