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, Aug 03 '12 at 08:45
The example looks right to me, the first FOO is surrounded by single quotes and must be skipped — daveoncode, Aug 03 '12 at 08:47
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, Aug 03 '12 at 08:52
no, only the literal "FOO" (with one space before and one after) should be replaced with "XXX" :P — daveoncode, Aug 03 '12 at 08:57
So you want to capture and replace *only* "FOO" strings that are *not* inside quotes? — Andrew Alcock, Aug 03 '12 at 09:01
This is a bit ambiguous: for example it depends on if you're reading left to right or right to left. If it's the latter then the 3rd FOO in from the right side shouldn't be replaced. — Adam Parkin, Aug 03 '12 at 18:24

MRAB · Accepted Answer · 2012-08-11T22:11:44.980

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)

Does not work for me, I get `' FOO '' def FOO ghi 345 '''' lmno 678 FOO '` as the output (the "XXX" are gone) — Adam Parkin, Aug 03 '12 at 18:20
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 — daveoncode, Aug 11 '12 at 21:02

davidrac · Answer 2 · 2012-08-03T09:03:59.830

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 03 '12 at 09:03

answered Aug 03 '12 at 08:58

davidrac

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Python's standard regex library 're' doesn't support variable-length lookbehinds, but there is an alternative regex library on PyPI which does at http://pypi.python.org/pypi/regex. – MRAB Aug 03 '12 at 18:20

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

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