Regex 使用正则表达式分别替换分隔符的开头和结尾

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"

s[1]“在1和之后填充一些文本”

我不是一个regex专业人士，也不知道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

为了记录在案，我完全不成功地提出了接近解决方案的想法：

# 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 $1" "$" 

#使用lookarounds我可以得到以下结果，但我需要它不是贪婪的
str_extract（s，perl）（（？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使用非贪婪运算符
？

\\$(.*?)\\$

或

哇。比我想象的简单！谢谢！你有没有试着简单地删除
$
？@casimirithippolyte。你的确切意思是什么？我想替换它们：）
# 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 $1" "$" 

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"

\\$(.*?)\\$

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