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r - 正则表达式在 R 中匹配具有相邻和非相邻单词重复的句子

问题描述

我有一个带有句子的数据框；在某些句子中，单词被多次使用：

df <- data.frame(Turn = c("well this is what the grumble about do n't they ?",
                          "it 's like being in a play-group , in n it ?",
                          "oh is that that steak i got the other night ?",
                          "well where have the middle sized soda stream bottle gone ?",
                          "this is a half day , right ? needs a full day",
                          "yourself , everybody 'd be changing your hair in n it ?",
                          "cos he finishes at four o'clock on that day anyway .",
                          "no no no i 'm dave and you 're alan .",
                          "yeah , i mean the the film was quite long though",
                          "it had steve martin in it , it 's a comedy",
                          "oh it is a dreary old day in n it ?",
                          "no it 's not mother theresa , it 's saint theresa .",
                          "oh have you seen that face lift job he wants ?",
                          "yeah bolshoi 's right so which one is it then ?"))

我想匹配那些单词，任何单词，重复一次或多次的句子。

编辑 1：

重复的单词**可以*相邻，但不必相邻。这就是为什么连续重复单词的正则表达式不能回答我的问题的原因。

我在这段代码中取得了适度的成功：

df[grepl("(\\w+\\b\\s)\\1{1,}", df$Turn),]
[1] well this is what the grumble about do n't they ?      
[2] it 's like being in a play-group , in n it ?           
[3] oh is that that steak i got the other night ?          
[4] this is a half day , right ? needs a full day          
[5] yourself , everybody 'd be changing your hair in n it ?
[6] no no no i 'm dave and you 're alan .                  
[7] yeah , i mean the the film was quite long though       
[8] it had steve martin in it , it 's a comedy             
[9] oh it is a dreary old day in n it ?

成功只是适度的，因为一些不应该匹配的句子被匹配，例如，，yourself , everybody 'd be changing your hair in n it ?而另一些不应该被匹配的句子，例如，no it 's not mother theresa , it 's saint theresa .。如何改进代码以产生完全匹配？

预期结果：

df
                                                         Turn
2                it 's like being in a play-group , in n it ?
3               oh is that that steak i got the other night ?
5               this is a half day , right ? needs a full day
8                       no no no i 'm dave and you 're alan .
9            yeah , i mean the the film was quite long though
10                 it had steve martin in it , it 's a comedy
11                        oh it is a dreary old day in n it ?
12        no it 's not mother theresa , it 's saint theresa .

编辑 2：

另一个问题是如何定义重复单词的确切数量。上述不完美的正则表达式匹配至少重复一次的单词。如果我将量词更改为{2}，从而寻找一个单词的三次出现，我会得到这个代码和这个结果：

df[grepl("(\\w+\\b\\s)\\1{2}", df$Turn),]
[1] no no no i 'm dave and you 're alan .         # "no" occurs 3 times

但同样，比赛并不完美，因为预期的结果是：

[1] no no no i 'm dave and you 're alan .          # "no" occurs 3 times
[2] it had steve martin in it , it 's a comedy     # "it" occurs 3 times

任何帮助深表感谢！

标签： rregextexttext-mining

解决方案

用于定义重复单词的确切数量的选项。

提取相同单词出现 3 次的句子

  1. 更改正则表达式。

    (\s?\b\w+\b\s)(.*\1){2}

    第 1 组捕获的 (\s?\b\w+\b\s)

    • \s? : 空格出现零次或一次。
    • \b\w+\b ：确切的单词字符。

    • \s ：空格出现一次。

      (.*\1) 被第 2 组捕获

      • (.*\1) ：在第 1 组再次匹配之前出现零次或多次的任何字符。

      • (.*\1){2} ：第 2 组匹配两次。

代码 

df$Turn[grepl("(\\s?\\b\\w+\\b\\s)(.*\\1){2}", df$Turn, perl = T)]
# [1] "no no no i 'm dave and you 're alan ."     
# [2] "it had steve martin in it , it 's a comedy"
  1. 将strsplit(split="\\s")句子拆分成单词。
    • 使用sapplyandtable统计每个列表元素中单词的出现次数，然后选择满足要求的句子。

代码 

library(magrittr)
df$Turn %<>% as.character()
s<-strsplit(df$Turn,"\\s") %>% sapply(.,function(i)table(i) %>% .[.==3])
df$Turn[which(s!=0)]
# [1] "no no no i 'm dave and you 're alan ."     
# [2] "it had steve martin in it , it 's a comedy"

希望这可以帮助你:)

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