Question

我想替换以下表达式中的$分隔符。

s <- "something before stuff $some text$ in between $1$ and after"

开头和结尾的替换必须是不同的，即

begin <- "<B>"     # replacement for 1st delimiter   
end <- "<E>"       # replacement for 2nd delimiter

结果应为

str_replace_all(s, SOME-REGEX-MAGIC)    
> [1] "something before stuff <B>some text<E> in between <B>1<E> and after"

我不是一名正则表达式专家，无法弄清楚如何分别处理分隔符的开头和结尾。

有什么想法吗？谢谢你的时间！

不成功的想法

仅仅为了记录我完全不成功的想法，以接近解决方案：

# Using lookarounds I get the following, but I would need it to be non-greedy
str_extract(s, perl("(?<=\\$).*(?=\\$)"))
"some text$ and some more $1"

# also greedy
str_match(s, "(\\$)(.*)(\\$)")
     [,1]                            [,2] [,3]                          [,4]
[1,] "$some text$ and some more $1$" "$"  "some text$ and some more $1" "$"

Answer 1

将此正则表达式与gsub()一起使用。替换使用反向引用（例如\\1）。

ptn <- "\\$(.*?)\\$" # Non-greedy find between delimiters
replacement <- "<B>\\1<E>"  # \\1 indicates back-reference
gsub(ptn, replacement, s)
[1] "something before stuff <B>some text<E> in between <B>1<E> and after"

后向引用\\1表示正则表达式中的第一个通配符表达式 - 即parens中的字符串 - (.*?)。 ?修饰符使匹配变得非贪婪。

Answer 2

使用非贪婪的算子?

\\$(.*?)\\$

或

\\$([^$]*)\\$

使用正则表达式单独替换分隔符的开头和结尾

2 个答案: