Expression in regular expression python

Question

I would like to make a regular expression for formatting a text, in which there can't be a { character except if it's coming with a backslash \ behind. The problem is that a backslash can escape itself, so I don't want to match \\{ for example, but I do want \\\{. So I want only an odd number of backslashs before a {. I can't just take it in a group and lookup the number of backslashs there are after like this:

s = r"a wei\\\{rd thing\\\\\{"
matchs = re.finditer(r"([^\{]|(\\+)\{)+", s)
for match in matchs:
    if len(match.group(2)) / 2 == len(match.group(2)) // 2: # check if it's even
        continue
    do_some_things()

Because the group 2 can be used more than one time, so I can access only to the last one (in this case, \\\\\) It would be really nice if we could just do something like "([^\{]|(\\+)(?if len(\2) / 2 == len(\2) // 2)\{)+" as regular expression, but, as far as I know, that is impossible. How can I do then ???

Beside the point, but to check if something's even, use modulo-2: 0 % 2 == 0, 1 % 2 == 1, 2 % 2 == 0, etc — wjandrea
– wjandrea, Commented May 9, 2020 at 18:56

wjandrea · Accepted Answer · 2020-05-10 00:58:45Z

1

This matches an odd number of backslashes followed by a brace:

(?<!\\)(\\\\)*(\\\{)

Breakdown:

(?<!\\) - Not preceded by a backslash, to accommodate the next bit
- This is called "negative lookbehind"
(\\\\)* - Zero or more pairs of backslashes
(\\\{) - A backslash then a brace

Matches:

\{
\\\{
\\\\\{

Non-matches:

\\{
\\\\{
\\\\\\{

Try it on RegExr

This was partly inspired by Vadim Baratashvili's answer

edited May 10, 2020 at 0:58

answered May 9, 2020 at 18:50

wjandrea

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3 Comments

Bananasmoothii Over a year ago

Thank you, but is there a way to not match the backslashs ? I would like to put everything in a lookbehind but Python don't want a variable-lenght text in a lookbehind. Also, this is a little bit dumb, because Python could just reverse the expression in the lookbehind ((abc)+ would become (cba)+) and do as always but backwards...

wjandrea Over a year ago

@Bananasmoothii Yes, just use a non-matching group, (?:\\\\)*

Bananasmoothii Over a year ago

I know that syntax, but I have a regex that looks like like \{([^\{%]+)(%[^\{%]+)?(?<!\\)(?:\\\\)*\} and I would like the first group to be blue and the second one to be green, but in "{a%b\\\\}", the backslashs stay black because they are either in the first either in the second group...

Vadim Baratashvili · Accepted Answer · 2020-05-09 19:06:05Z

0

I think you can use this as solution: ([^\\](\\\\){0,})(\{)

We can check that between the last character that is not a backslash there are 0 or more pairs of backslashes and then goes {if part of the text matches the pattern, then we can replace it with the first group $1 (a character that is not a slash plus 0 or more pairs of slashes), so we will find and replace not escaped { .

If we want to find escaped { we ca use this expression: ([^\\](\\\\){0,})(\\\{) - second group of match is \{

edited May 9, 2020 at 19:06

answered May 9, 2020 at 17:39

Vadim Baratashvili

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6 Comments

wjandrea Over a year ago

Awesome. I just rearranged it to make it clearer. I can't get it to work though. It doesn't find any matches in the example.

Vadim Baratashvili Over a year ago

@wjandrea if you add { with paired backslashes or without backslashes to the text, { will be replaced by an empty string and finded if we do not need to replace anything (then we have to process it as we need)

wjandrea Over a year ago

I think that's the opposite of what OP wants, but the question is not totally clear. I posted my own answer partly inspired by yours though, thanks!

Vadim Baratashvili Over a year ago

@wjandrea or you can use ([^\\](\\\\){0,})(\\\{) to find \{ \\\{ \\\\\{

Bananasmoothii Over a year ago

Why do you use {0,} instead of * ?

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