Python non-greedy regexes
Question:
How do I make a python regex like "(.*)" such that, given "a (b) c (d) e" python matches "b" instead of "b) c (d"?
I know that I can use "[^)]" instead of ".", but I’m looking for a more general solution that keeps my regex a little cleaner. Is there any way to tell python “hey, match this as soon as possible”?
Answers:
You seek the all-powerful *?
From the docs, Greedy versus Non-Greedy
the non-greedy qualifiers *?, +?, ??, or {m,n}? […] match as little
text as possible.
Would not \(.*?\) work? That is the non-greedy syntax.
>>> x = "a (b) c (d) e"
>>> re.search(r"(.*)", x).group()
'(b) c (d)'
>>> re.search(r"(.*?)", x).group()
'(b)'
The ‘*‘, ‘+‘, and ‘?‘ qualifiers are all greedy; they match as much text as possible. Sometimes this behavior isn’t desired; if the RE <.*> is matched against ‘<H1>title</H1>‘, it will match the entire string, and not just ‘<H1>‘. Adding ‘?‘ after the qualifier makes it perform the match in non-greedy or minimal fashion; as few characters as possible will be matched. Using .*? in the previous expression will match only ‘<H1>‘.
Do you want it to match “(b)”? Do as Zitrax and Paolo have suggested. Do you want it to match “b”? Do
>>> x = "a (b) c (d) e"
>>> re.search(r"((.*?))", x).group(1)
'b'
Using an ungreedy match is a good start, but I’d also suggest that you reconsider any use of .* — what about this?
groups = re.search(r"([^)]*)", x)
As the others have said using the ? modifier on the * quantifier will solve your immediate problem, but be careful, you are starting to stray into areas where regexes stop working and you need a parser instead. For instance, the string “(foo (bar)) baz” will cause you problems.
To start with, I do not suggest using “*” in regexes. Yes, I know, it is the most used multi-character delimiter, but it is nevertheless a bad idea. This is because, while it does match any amount of repetition for that character, “any” includes 0, which is usually something you want to throw a syntax error for, not accept. Instead, I suggest using the + sign, which matches any repetition of length > 1. What’s more, from what I can see, you are dealing with fixed-length parenthesized expressions. As a result, you can probably use the {x, y} syntax to specifically specify the desired length.
However, if you really do need non-greedy repetition, I suggest consulting the all-powerful ?. This, when placed after at the end of any regex repetition specifier, will force that part of the regex to find the least amount of text possible.
That being said, I would be very careful with the ? as it, like the Sonic Screwdriver in Dr. Who, has a tendency to do, how should I put it, “slightly” undesired things if not carefully calibrated. For example, to use your example input, it would identify ((1) (note the lack of a second rparen) as a match.
How do I make a python regex like "(.*)" such that, given "a (b) c (d) e" python matches "b" instead of "b) c (d"?
I know that I can use "[^)]" instead of ".", but I’m looking for a more general solution that keeps my regex a little cleaner. Is there any way to tell python “hey, match this as soon as possible”?
You seek the all-powerful *?
From the docs, Greedy versus Non-Greedy
the non-greedy qualifiers
*?,+?,??, or{m,n}?[…] match as little
text as possible.
Would not \(.*?\) work? That is the non-greedy syntax.
>>> x = "a (b) c (d) e"
>>> re.search(r"(.*)", x).group()
'(b) c (d)'
>>> re.search(r"(.*?)", x).group()
'(b)'
The ‘
*‘, ‘+‘, and ‘?‘ qualifiers are all greedy; they match as much text as possible. Sometimes this behavior isn’t desired; if the RE<.*>is matched against ‘<H1>title</H1>‘, it will match the entire string, and not just ‘<H1>‘. Adding ‘?‘ after the qualifier makes it perform the match in non-greedy or minimal fashion; as few characters as possible will be matched. Using.*?in the previous expression will match only ‘<H1>‘.
Do you want it to match “(b)”? Do as Zitrax and Paolo have suggested. Do you want it to match “b”? Do
>>> x = "a (b) c (d) e"
>>> re.search(r"((.*?))", x).group(1)
'b'
Using an ungreedy match is a good start, but I’d also suggest that you reconsider any use of .* — what about this?
groups = re.search(r"([^)]*)", x)
As the others have said using the ? modifier on the * quantifier will solve your immediate problem, but be careful, you are starting to stray into areas where regexes stop working and you need a parser instead. For instance, the string “(foo (bar)) baz” will cause you problems.
To start with, I do not suggest using “*” in regexes. Yes, I know, it is the most used multi-character delimiter, but it is nevertheless a bad idea. This is because, while it does match any amount of repetition for that character, “any” includes 0, which is usually something you want to throw a syntax error for, not accept. Instead, I suggest using the + sign, which matches any repetition of length > 1. What’s more, from what I can see, you are dealing with fixed-length parenthesized expressions. As a result, you can probably use the {x, y} syntax to specifically specify the desired length.
However, if you really do need non-greedy repetition, I suggest consulting the all-powerful ?. This, when placed after at the end of any regex repetition specifier, will force that part of the regex to find the least amount of text possible.
That being said, I would be very careful with the ? as it, like the Sonic Screwdriver in Dr. Who, has a tendency to do, how should I put it, “slightly” undesired things if not carefully calibrated. For example, to use your example input, it would identify ((1) (note the lack of a second rparen) as a match.