正則表達式； 刪除所有標點符號，除了

Question

我有以下正則表達式可以在任何空格或標點符號上拆分。 如何從:punct:排除 1 個或多個標點符號？ 假設我想排除撇號和逗號。 我知道我可以明確使用[all punctuation marks in here]而不是[[:punct:]]但我希望有一種排除方法。

X <- "I'm not that good at regex yet, but am getting better!"
strsplit(X, "[[:space:]]|(?=[[:punct:]])", perl=TRUE)

 [1] "I"       "'"       "m"       "not"     "that"    "good"    "at"      "regex"   "yet"    
[10] ","       ""        "but"     "am"      "getting" "better"  "!"

Answer 1

我不清楚你想要的結果是什么，但你可以使用像這個答案這樣的否定類。

R> strsplit(X, "[[:space:]]|(?=[^,'[:^punct:]])", perl=TRUE)[[1]]
 [1] "I'm"     "not"     "that"    "good"    "at"      "regex"   "yet,"   
 [8] "but"     "am"      "getting" "better"  "!"    

Answer 2

如果右側的下一個字符是'或,則您可以直接使用(?![',])負前瞻對 PCRE 子模式施加限制，該負前瞻會導致匹配失敗：

[[:space:]]|(?=(?![',])[[:punct:]])
               ^^^^^^^^ 

請參閱正則表達式演示。

細節

• [[:space:]] - 任何空格
• | - 要么
• (?=(?![',])[[:punct:]]) - 一個正向前瞻，要求在當前位置的右側，沒有' and ,並且有任何 1 個標點字符那不是'或, （實際上，需要除'和,之外'任何標點符號）。

查看R 在線演示

X <- "I'm not that good at regex yet, but am getting better!"
strsplit(X, "[[:space:]]|(?=(?![',])[[:punct:]])", perl=TRUE)
[[1]]
 [1] "I'm"     "not"     "that"    "good"    "at"      "regex"   "yet,"   
 [8] "but"     "am"      "getting" "better"  "!"