在 LIKE 上合并两个表，但对于整个字符串而不是字符串的一部分答案

【问题标题】：Merge two tables on LIKE but for whole string not parts of strings在 LIKE 上合并两个表，但对于整个字符串而不是字符串的一部分
【发布时间】：2017-06-23 19:42:48
【问题描述】：

这是我的第一篇文章/问题，请善待。我有一个这样的数据框：

        id                             product
1 00109290                    Wax Salt; Pepper
2 23243242                          Wood Stuff
3 23242433   Magic Unicorn Powder and My Tears
4 23778899                             gelatin
5 25887766                                tin;
6  7786655             fart noises, and things
7  3432422 --spearmint bacon& hydrangia leaves

我有一个这样的查找表：

        ingredients
1               wax
2              salt
3              wood
4          my tears
5    unicorn powder
6           gelatin
7               tin
8  hydrangia leaves
9         spearmint
10            bacon

我想将它们合并到整个字符串上，所以我得到了这个：

     id                             product      ingredients
1  00109290                    Wax Salt; Pepper              wax
2  00109290                    Wax Salt; Pepper             salt
3  23243242                          Wood Stuff             wood
4  23242433   Magic Unicorn Powder and My Tears         my tears
5  23242433   Magic Unicorn Powder and My Tears   unicorn powder
6  23778899                             gelatin          gelatin
7  25887766                                tin;              tin
8   3432422 --spearmint bacon& hydrangia leaves hydrangia leaves
9   3432422 --spearmint bacon& hydrangia leaves        spearmint
10  3432422 --spearmint bacon& hydrangia leaves            bacon

相反，我得到了这个（注意第 7 行不需要）：

         id                             product      ingredients
1  00109290                    Wax Salt; Pepper              wax
2  00109290                    Wax Salt; Pepper             salt
3  23243242                          Wood Stuff             wood
4  23242433   Magic Unicorn Powder and My Tears         my tears
5  23242433   Magic Unicorn Powder and My Tears   unicorn powder
6  23778899                             gelatin          gelatin
7  23778899                             gelatin              tin
8  25887766                                tin;              tin
9   3432422 --spearmint bacon& hydrangia leaves hydrangia leaves
10  3432422 --spearmint bacon& hydrangia leaves        spearmint
11  3432422 --spearmint bacon& hydrangia leaves            bacon

我非常接近，但我错误地将“明胶”与“锡”匹配。我想匹配整个单词，而不是单词的一部分。我尝试了许多不同的技术，最接近的是：

library(sqldf)
id <- c('00109290', '23243242', '23242433', 
        '23778899', '25887766', '7786655', 
        '3432422')
product <- c('Wax Salt; Pepper', 'Wood Stuff', 
             'Magic Unicorn Powder and My Tears', 
             'gelatin', 'tin;', 'fart noises, and things', 
             '--spearmint bacon& hydrangia leaves')

ingredients <- c('wax', 'salt', 'wood', 'my tears', 
                 'unicorn powder', 'gelatin', 'tin', 
                 'hydrangia leaves', 
                 'spearmint', 'bacon') 

products <- data.frame(id, product)
ingred <- data.frame(ingredients)    
new_df <- sqldf("SELECT * from products 
                 join ingred on product LIKE '%' || ingredients || '%'")

非常感谢任何建议。也许需要一种完全不同的方法？我也欢迎就问题的质量提出建议，这是我的第一次，所以你最好马上给我安排。

【问题讨论】：

对你来说一个完整的词是什么？空格之间的一大堆字符？
非常好的问题。因为如果它是空格之间的字符，那么我不会将 'salt' 匹配到 'salt;'，而我的数据集具有这种类型的东西。我不想在一个词中匹配。所以'salt'应该匹配'salt;'但不是 'salty' 和 'tin' 应该匹配 'tin' 或 ';tin' 但不是 'gelatin'。有了你的一个问题，我意识到......这很可能是正则表达式的事情，我认为我不能使用 sqldf。谢谢！
我检查了docs，但找不到任何提及它支持正则表达式的信息。在 MySQL 中，存在REGEXP 运算符，并使用[[:<:]] 和[[:>:]] 标记单词边界。在 PostgreSQL 中，使用\y。在正则表达式中，\b 标记单词边界。不幸的是，我认为 sqldf 不可行。
@WiktorStribiżew 我认为通过使用 LIKE 保留现有解决方案，并使用正则表达式添加一个步骤以过滤出错误匹配项是否可行？
你能否列出一个不属于单词的非空格事物的简短列表？您的示例输入似乎只包含“;”在那个名单上。我肯定会添加“，”和“&”。

标签： r regex sql-like sqldf

【解决方案1】：

使用fuzzyjoin包的解决方案，以及来自stringr的str_detect：

library(fuzzyjoin)
library(stringr)

f <- function(x, y) {
  # tests whether y is an ingredient of x
  str_detect(x, regex(paste0("\\b", y, "\\b"), ignore_case = TRUE))
}

fuzzy_join(products, 
           ingred, 
           by = c("product" = "ingredients"), 
           match_fun = f)
#         id                           product    ingredients
# 1   109290                  Wax Salt; Pepper            wax
# 2   109290                  Wax Salt; Pepper           salt
# 3 23243242                        Wood Stuff           wood
# 4 23242433 Magic Unicorn Powder and My Tears       my tears
# 5 23242433 Magic Unicorn Powder and My Tears unicorn powder
# 6 23778899                           gelatin        gelatin

数据

products <- read.table(text = "
        id                             product
1 00109290                  'Wax Salt; Pepper'
2 23243242                        'Wood Stuff'
3 23242433 'Magic Unicorn Powder and My Tears'
4 23778899                             gelatin                  
  ", stringsAsFactors = FALSE)

ingred <- read.table(text = "
       ingredients
1              wax
2             salt
3             wood
4       'my tears'
5 'unicorn powder'
6          gelatin
7              tin
  ", stringsAsFactors = FALSE)

【讨论】：

【解决方案2】：

考虑为关键字前后的一个空格添加OR 条件，然后完全匹配并替换任何特殊字符/标点符号。

new_df <- sqldf("SELECT * from products 
                 join ingred on Replace(product, ';', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, ';', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, ';', '') = ingredients
                ")

您甚至可以UNION 获取不同的特殊字符。下面的例子替换了分号和感叹号：

new_df <- sqldf("SELECT * from products 
                 join ingred on Replace(product, ';', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, ';', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, ';', '') = ingredients
                 UNION    
                 SELECT * from products 
                 join ingred on Replace(product, '!', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '!', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '!', '') = ingredients
                ")

对于许多 UNIONs，考虑让 R 连接 SQL 语句：

sql <- paste(lapply(c("!", "#", "$", "%", "(", ")", ":", ";", ".", "?", ">", "<", "/", "\\\\", "|"), 
             function(i)
                paste0("SELECT * from products 
                        join ingred on Replace(product, '", i, "', '') LIKE '% ' || ingredients || '%'
                                    OR Replace(product, '", i, "', '') LIKE '%' || ingredients || ' %'
                                    OR Replace(product, '", i, "', '') = ingredients
                       ")
       ), collapse = "UNION ")

cat(paste(sql))

SELECT * from products 
                   join ingred on Replace(product, '!', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '!', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '!', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '#', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '#', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '#', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '$', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '$', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '$', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '%', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '%', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '%', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '(', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '(', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '(', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, ')', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, ')', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, ')', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, ':', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, ':', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, ':', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, ';', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, ';', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, ';', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '.', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '.', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '.', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '?', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '?', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '?', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '>', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '>', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '>', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '<', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '<', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '<', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '/', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '/', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '/', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '\\', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '\\', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '\\', '') = ingredients
                   UNION SELECT * from products 
                   join ingred on Replace(product, '|', '') LIKE '% ' || ingredients || '%'
                             OR Replace(product, '|', '') LIKE '%' || ingredients || ' %'
                             OR Replace(product, '|', '') = ingredients

【讨论】：

天哪……我怎么没想到呢？我的第一个帖子/问题，现在我很尴尬。非常感谢。
我认为“gelatin”就像“%tin %”（在第一个和第二个“OR”之间）。而且“明胶和盐”也会有问题。
@DieselBlue 我实际上在我的第一条评论中问过这个问题。如果你没有提到你需要在tin; 中匹配tin，我会发布这个答案。
@Wiktor Stribiżew ...你是对的，这不会匹配 'tin;'和 dziękuję za pomoc。
@DieselBlue ... 查看更新，您可以在其中使用 Replace 删除特殊字符，例如 ;。