From c02037d62284f4d4ca6b22f2ed05165ce2014951 Mon Sep 17 00:00:00 2001
From: "Miss Islington (bot)"
 <31488909+miss-islington@users.noreply.github.com>
Date: Fri, 24 Nov 2017 21:01:39 -0800
Subject: [PATCH] bpo-30004: Fix the code example of using group in Regex Howto
 Docs (GH-4443) (GH-4555)

The provided code example was supposed to find repeated words, however it returned false results.
(cherry picked from commit 610e5afdcbe3eca906ef32f4e0364e20e1b1ad23)
---
 Doc/howto/regex.rst | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/Doc/howto/regex.rst b/Doc/howto/regex.rst
index 0d0e9f56093..082fc012991 100644
--- a/Doc/howto/regex.rst
+++ b/Doc/howto/regex.rst
@@ -840,7 +840,7 @@ backreferences in a RE.
 
 For example, the following RE detects doubled words in a string. ::
 
-   >>> p = re.compile(r'(\b\w+)\s+\1')
+   >>> p = re.compile(r'\b(\w+)\s+\1\b')
    >>> p.search('Paris in the the spring').group()
    'the the'
 
@@ -947,9 +947,9 @@ number of the group.  There's naturally a variant that uses the group name
 instead of the number. This is another Python extension: ``(?P=name)`` indicates
 that the contents of the group called *name* should again be matched at the
 current point.  The regular expression for finding doubled words,
-``(\b\w+)\s+\1`` can also be written as ``(?P<word>\b\w+)\s+(?P=word)``::
+``\b(\w+)\s+\1\b`` can also be written as ``\b(?P<word>\w+)\s+(?P=word)\b``::
 
-   >>> p = re.compile(r'(?P<word>\b\w+)\s+(?P=word)')
+   >>> p = re.compile(r'\b(?P<word>\w+)\s+(?P=word)\b')
    >>> p.search('Paris in the the spring').group()
    'the the'