Find and replace with two strings between it with regular expression

Question

In Python, I tried to replace two strings in between a regular expression match.

import re

a = "("
b = ")"
string = "foo bar foo foo bar bar foofoofoo foo foo"

regex = "(foo(.[foo]{1,}))|foo"
print(re.sub(regex, a + string + b, string))

What I thought was going to print:

(foo) bar (foo foo) bar bar (foofoofoo) (foo foo)

What it actually printed:

(foo bar foo foo bar bar foofoofoo foo foo) bar (foo bar foo foo bar bar foofoofoo foo foo) bar bar (foo bar foo foo bar bar foofoofoo foo foo) (foo bar foo foo bar bar foofoofoo foo foo)

Should I use loops or is there a function for that?

Dave F · Accepted Answer · 2016-01-09 19:47:31Z

2

re.sub is defined as

re.sub(pattern , <what to replace "pattern" with>, input_string)

Your code is asking re.sub to replace your matched pattern with your input_string surrounded by parenthesis which is not what you want.

If you just want to replace all foo's with parenthesis, try

regex = r"((foo){2,}|foo(\s+foo)*)"
a = "("
b = ")"
input_string = "foo bar foo foo bar bar foofoofoo foo foo"
print(re.sub(regex, a + r'\1' + b, input_string))

Output:

(foo) bar (foo foo) bar bar (foofoofoo) (foo foo)

edited Jan 9, 2016 at 19:47

Dave F

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answered Jan 9, 2016 at 18:52

Martin Konecny

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6 Comments

Anastasia Dunbar Over a year ago

What if I wanted to change the first string and second string to close with user input? It prints () bar () () bar bar () () () when I use variable strings.

Anastasia Dunbar Over a year ago

Never mind it's a+r'\1'+b sorry. I didn't know what r really does. But I guess when it's for when it's regular expression or it's not.

Anastasia Dunbar Over a year ago

How do I do so user's regex string input with r at the beginning? It doesn't do anything without it.

Martin Konecny Over a year ago

If you don't put the r, then make sure you use double backslash instead of single: \\1

Anastasia Dunbar Over a year ago

I am not talking about without the r in re.sub 's replacement. I am saying that re.sub 's pattern always work when a r is before a string. But what do I do if an user inputs pattern but there isn't a r before it because you can't do 'r'+str(raw_input("Input pattern: ")).

|

Dave F · Accepted Answer · 2016-01-09 19:42:37Z

2

You're not using a back reference to your original match. Instead you're replacing the match with the original string. This is why your string is getting longer. You need to use \1 to refer to the pattern matched in the first set of brackets.

I used the following code and got the output that you wanted:

print(re.sub(r"((foo){2,}|foo(\s+foo)*)",r'(\1)',string))

EDIT: I don't have the reputation to comment on the answer marked as correct (my account was reset after being dormant). However, the output is wrong based on the original question.

I got the output:

(foo) bar (foo foo) bar bar (foofoofoo) (foo foo)

EDIT: I corrected the original answer. I didn't think that I could with my reputation.

edited Jan 9, 2016 at 19:42

answered Jan 9, 2016 at 19:30

Dave F

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1 Comment

Wiktor Stribiżew Over a year ago

You can always correct what needs correction, and if it is helpful and correct, any reputation is enough :)

Dan Lowe · Accepted Answer · 2016-01-09 18:52:31Z

1

You need to read the documentation for re.sub again. This is its definition:

re.sub(pattern, repl, string, count=0, flags=0)

You are calling:

re.sub(regex, a + string + b, string)

No matter what matches, you are replacing the match with your entire original string surrounded by parens.

Your regex as written has two capture groups (defined by parens). You can refer to them inside the replacement string as \1 or \2. You need to replace a+string+b with something else that will print what you are looking for instead of the entire string.

answered Jan 9, 2016 at 18:52

Dan Lowe

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