Can anyone tell me what does \"\\1\" mean in the following regular expression in Python?
re.sub(r\'(\\b[a-z]+) \\1\', r\'\\1\', \'cat in the the hat\')
Example
The following code using Python regex to find the repeating digits in given string
import re
result = re.search(r'(\d)\1{3}','54222267890' )
print result.group()
This gives the output
2222
The first \1
means the first group - i.e. the first bracketed expression (\b[a-z]+)
From the docs \number
"Matches the contents of the group of the same number. Groups are numbered starting from 1. For example, (.+) \1 matches 'the the' or '55 55', but not 'thethe' (note the space after the group)"
In your case it is looking for a repeated "word" (well, block of lower case letters).
The second \1
is the replacement to use in case of a match, so a repeated word will be replaced by a single word.
\1
is a backreference.
It matches, what ever matched in your brackets, in this case the
You are basically saying
cat in (' ''the')' the' hat
From the python docs for the re module:
\number
Matches the contents of the group of the same number. Groups are numbered starting from 1. For example,
(.+) \1
matches'the the'
or'55 55'
, but not'thethe'
(note the space after the group). This special sequence can only be used to match one of the first 99 groups. If the first digit of number is 0, or number is 3 octal digits long, it will not be interpreted as a group match, but as the character with octal value number. Inside the'['
and']'
of a character class, all numeric escapes are treated as characters.
Your example is basically the same as what is explained in the docs.
\1
is equivalent to re.search(...).group(1)
, the first parentheses-delimited expression inside of the regex.
It's also, fun fact, part of the reason that regular expressions are significantly slower in Python and other programming languages than required to be by CS theory.