"?!" in regular expression

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import re
a = re.compile('^a(?!abc)')
b = re.compile('^a(?!bac)$')
c = re.compile('^a((?!ba).)*$')

a1 = "abbc"
print(a.search(a1)) # <re.Match object; span=(0, 1), match='a'>
print(b.search(a1)) # None
print(c.search(a1)) # <re.Match object; span=(0, 4), match='abbc'>

I don't understand the expression "?!" in python Regular expression.

Why is only a searched in the first case?

I wonder if all the characters of abbc can be searched because it is a match that the string abc does not come next.

Why is b none?

Doesn't it all apply to abbc since it starts with a and doesn't end with bac?

python-re python
1 Answers

I don't understand the expression "?!" in python Regular expression

Well, ?! is not an expression, regular or otherwise.

The documentation says:

  • (?=...)

    Matches if ... matches next, but doesn’t consume any of the string. This is called a lookahead assertion. For example, Isaac (?=Asimov) will match 'Isaac ' only if it’s followed by 'Asimov'.

  • (?!...)

    Matches if ... doesn’t match next. This is a negative lookahead assertion. For example, Isaac (?!Asimov) will match 'Isaac ' only if it’s not followed by 'Asimov'.

(including the definition of a positive lookahead assertion for comparison).

So, the expression is really (?!...).

Note that the (? sequence is used for several extensions to plain regular expressions, and they're all documented.

Why is only a searched in the first case?

Because a is the only character the pattern asks for. The lookahead group could prevent a from matching (if it was followed by abc), but it wouldn't be part of the match either way.

user_2128580
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