regex - 如何匹配非连续重复n次以上的模式？

Question

我正在尝试在此示例向量上匹配模式“城市，州”(例如“奥斯汀，德克萨斯州”)

> s <- c("Austin, TX", "Forth Worth, TX", "Ft. Worth, TX", 
"Austin TX", "Austin, TX, USA", "Ft. Worth, TX, USA")

> grepl('[[:alnum:]], [[:alnum:]$]', s)
[1]  TRUE  TRUE  TRUE FALSE  TRUE  TRUE

但是，有两种情况我想检索 FALSE:

-当超过 1 个逗号时(即 "Austin, TX, USA")

-当逗号之前有另一个标点符号时(即“Ft. Worth, TX”)

Answer 1

您可以使用以下正则表达式模式:

grepl("^[a-z ]+, [a-z]+$", subject, perl=TRUE, ignore.case=TRUE);

Regex101 Demo

正则表达式说明:

^[a-z ]+, [a-z]+$/gmi

    ^ assert position at start of a line
    [a-z ]+ match a single character present in the list below
        Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
        a-z a single character in the range between a and z (case insensitive)
         the literal character  
    , matches the characters , literally
    [a-z]+ match a single character present in the list below
        Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
        a-z a single character in the range between a and z (case insensitive)
    $ assert position at end of a line
    ignore.case: insensitive. Case insensitive match (ignores case of [a-zA-Z])