不包括模式前的一组单词。

Question

我想抓住所有提到的“养老金”(资本不敏感，包括养老金，养老金领取者，但排除不相关的词，如“停职”。然而，我想把养老金排除在“工作部”之前，但我无法抓住整个说法。到目前为止，我已经：

sentences <- c("department of work and pensions", "and pensioners", "pensioners", "Pensions", "suspension")
try <- grepl("(?<!department of work and )^pension*", ignore.case = T, perl = T, sentences)
try

有什么建议吗？

Answer 1

我们可以用

grepl("\\bpension\\S+", sentences, ignore.case = TRUE) & 
      !grepl("department of work .*\\bpension\\S+", sentences, ignore.case = TRUE)

Answer 2

grep('(?<!department of work and )\\bpension', sentences, 
        value = TRUE, ignore.case = TRUE, perl = TRUE)

[1] "and pensioners" "pensioners"     "Pensions" 

Answer 3

您可以使用单个模式，该模式将说明单词之间的任何空格，并且只在单词边界匹配pension：

sentences <- c("department of work and pensions", "and pensioners", "pensioners", "Pensions", "suspension")
grepl("\\bdepartment of work and \\w+(*SKIP)(*F)|\\bpension", ignore.case = T, perl = T, sentences)
## => [1] FALSE  TRUE  TRUE  TRUE FALSE

请参阅R演示和regex演示。

详细信息

• \bdepartment of work and \w+ -单词边界\b，department of work and +空格+一个或多个单词字符
• (*SKIP)(*F) -省略到目前为止匹配的所有文本，然后从失败位置开始下一次匹配搜索
• | -或
• \bpension -单词边界\b和pension子字符串。