正则表达式查找并替换多个

Question

I am trying to write a regular expression that will match all cases of

[[any text or char her]]

in a series of text.

Eg:

My name is [[Sean]]
There is a [[new and cool]] thing here.

This all works fine using my regex.

data = "this is my tes string [[ that does some matching ]] then returns."
p = re.compile("\[\[(.*)\]\]")
data = p.sub('STAR', data)

The problem is when I have multiple instances of the match occuring :[[hello]] and [[bye]]

Eg:

data = "this is my new string it contains [[hello]] and [[bye]] and nothing else"
p = re.compile("\[\[(.*)\]\]")
data = p.sub('STAR', data)

This will match the opening bracket of hello and the closing bracket of bye. I want it to replace them both.

Answer 1

.* is greedy and matches as much text as it can, including ]] and [[ , so it plows on through your "tag" boundaries.

A quick solution is to make the star lazy by adding a ? :

p = re.compile(r"\[\[(.*?)\]\]")

A better (more robust and explicit but slightly slower) solution is to make it clear that we cannot match across tag boundaries:

p = re.compile(r"\[\[((?:(?!\]\]).)*)\]\]")

Explanation:

\[\[        # Match [[
(           # Match and capture...
 (?:        # ...the following regex:
  (?!\]\])  # (only if we're not at the start of the sequence ]]
  .         # any character
 )*         # Repeat any number of times
)           # End of capturing group
\]\]        # Match ]]

Answer 2

Use ungreedy matching .*? <~~ the ? after a + or * makes it match as few characters as possible. The default is to be greedy, and consume as many characters as possible.

p = re.compile("\[\[(.*?)\]\]")

Answer 3

You can use this:

p = re.compile(r"\[\[[^\]]+\]\]")

>>> data = "this is my new string it contains [[hello]] and [[bye]] and nothing else"
>>> p = re.compile(r"\[\[[^\]]+\]\]")
>>> data = p.sub('STAR', data)
>>> data
'this is my new string it contains STAR and STAR and nothing else'