Python regex to replace a string if not surrounded by single quotes

Question

I'm trying to write a regex to replace strings if not surrounded by single quotes. For example I want to replace FOO with XXX in the following string:

string = "' FOO ' abc 123 ' def FOO ghi 345 ' FOO '' FOO ' lmno 678 FOO '"

the desired output is:

output = "' FOO ' abc 123 ' def FOO ghi 345 ' XXX '' XXX ' lmno 678 FOO '"

My current regex is:

myregex = re.compile("(?<!')+( FOO )(?!')+", re.IGNORECASE)

I think I have to use look-around operators, but I don't understand how... regex are too complicated to me :D

Can you help me?

I think your example is broken. Why isn't the first " abc 123 " not replaced with XXX? — Andrew Alcock
– Andrew Alcock, Commented Aug 3, 2012 at 8:45
The example looks right to me, the first FOO is surrounded by single quotes and must be skipped — daveoncode
– daveoncode, Commented Aug 3, 2012 at 8:47
Agree with you on the first FOO. However, doesn't that mean that the bit starting abc is /outside/? If so, the result should be: "' FOO ' XXX ' def FOO ghi 345 ' XXX '' XXX ' lmno 678 FOO '". Correct? — Andrew Alcock
– Andrew Alcock, Commented Aug 3, 2012 at 8:52
no, only the literal "FOO" (with one space before and one after) should be replaced with "XXX" :P — daveoncode
– daveoncode, Commented Aug 3, 2012 at 8:57

MRAB · Accepted Answer · 2012-08-11 22:11:44Z

3

Here's how it could be done:

import re

def replace_FOO(m):
    if m.group(1) is None:
        return m.group()

    return m.group().replace("FOO", "XXX")

string = "' FOO ' abc 123 ' def FOO ghi 345 ' FOO '' FOO ' lmno 678 FOO '"

output = re.sub(r"'[^']*'|([^']*)", replace_FOO, string)

print(string)
print(output)

[EDIT]

The re.sub function will accept as a replacement either a string template or a function. If the replacement is a function, every time it finds a match it'll call the function, passing the match object, and then use the returned value (which must be a string) as the replacement string.

As for the pattern itself, as it searches, if there's a ' at the current position it'll match up to and including the next ', otherwise it'll match up to but excluding the next ' or the end of the string.

The replacement function will be called on each match and return the appropriate result.

Actually, now I think about it, I don't need to use a group at all. I could do this instead:

def replace_FOO(m):
    if m.group().startswith("'"):
        return m.group().replace("FOO", "XXX")

    return m.group()

string = "' FOO ' abc 123 ' def FOO ghi 345 ' FOO '' FOO ' lmno 678 FOO '"

output = re.sub(r"'[^']*'|[^']+", replace_FOO, string)

edited Aug 11, 2012 at 22:11

answered Aug 3, 2012 at 18:16

MRAB

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2 Comments

Adam Parkin Over a year ago

Does not work for me, I get ' FOO '' def FOO ghi 345 '''' lmno 678 FOO ' as the output (the "XXX" are gone)

daveoncode Over a year ago

It works as expected for me (Python 2.7.1) thanks a lot! It would be very useful if you could explain the code, since I'm a Python and regex newbie :P

davidrac · Accepted Answer · 2012-08-03 09:03:59Z

2

This is hard to do without variable length lookbehind. I'm not sure if python regex support it. Anyway, a simple solution is the following:

Use this regex: (?:[^'\s]\s*)(FOO)(?:\s*[^'\s])

The first capture group should return the right result.

In case this is always a quote with a single space after it, as in your example, you can use fixed length lookbehind: (?<=[^'\s]\ )FOO(?=\s*[^'\s]) which will match exactly the one you want.

edited Aug 3, 2012 at 9:03

answered Aug 3, 2012 at 8:58

davidrac

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1 Comment

MRAB Over a year ago

Python's standard regex library 're' doesn't support variable-length lookbehinds, but there is an alternative regex library on PyPI which does at pypi.python.org/pypi/regex.

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