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 😀

Can you help me?

Answer 1

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)