Question

我有以下正则表达式，可以分割任何空格或标点符号。如何从:punct:中排除1个或多个标点字符？假设我想排除撇号和逗号。我知道我可以明确使用[all punctuation marks in here]代替[[:punct:]]，但我希望有一种排除方法。

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

我不清楚你想要的结果是什么，但你可以使用负面的like this answer。

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

Answer 2

如果右侧的下一个字符为(?![',])或'，则您可以直接使用, negative lookahead对PCRE子模式施加限制，但匹配失败：< / p>

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

请参阅regex demo。

<强>详情

[[:space:]] - 任何空白
| - 或
(?=(?![',])[[:punct:]]) - 一个积极的前瞻，要求在当前位置的右侧，没有'和,并且有任何1个标点符号不是'或,（实际上，需要'和,以外的任何标点符号。

请参阅R online demo

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"  "!"

正则表达式;消除所有标点符号除外

2 个答案: