正则表达式：如何根据负前瞻和后视将子字符串放在括号内？

Question

my goal is to put a substring within parenthesis according to some specific rules.

For example, here is a text:

text = 'cake OR ice cream'

And my goal is to transform this original text into this:

'cake OR (ice AND cream)'

As you can see, the ultimate goal is to preserve some Boolean logics within the text.

First step is to add the TO_PARENTHESIS that we will use as some sort of anchor. I can do it using negative lookbehind & lookahead:

import regex as re

text = re.sub(r'(?<!OR|AND|NOT|\(|\)|")(\s+)(?!OR|AND|NOT|\(|\)|")', r' TO_PARENTHESIS ', text)

This will find empty spaces & replace those with TO_PARENTHESIS but only the ones that are between two non-boolean keywords (and non-special characters).

Here is what we get:

cake OR ice TO_PARENTHESIS cream

Now, my question is, how do I put parenthesis at the very specific points, to have something like this:

cake OR (ice TO_PARENTHESIS cream)

I tried: (?<!OR|AND|NOT|\\(|\\)|")(.*TO_PARENTHESIS.*)(?!OR|AND|NOT|\\(|\\)|") but this will select the entire text & not just the ice TO_PARENTHESIS cream as expected.

So two questions:

• How to select the correct group?
• How to replace that selected group by itself plus the two parenthesis?

And last step would be to replace the TO_PARENTHESIS with AND to finally get our 'cake OR (ice AND cream)' .

Answer 1

Maybe...

import re

# starting string
text = 'cake OR ice cream'

# first pattern that finds the space between two lowercase letters (assuming it's always OR, AND, NOT, etc.)
pattern_1 = re.compile(r'([a-z])\s([a-z])') #replace with \1 AND \2

# capture the word (via a boundary) before the AND and the word after the AND
pattern_2 = re.compile(r'(\b\w+\b\sAND\s\b\w+)') #replace with (\1)

# show the starting text
print(text)

# make 'cake OR ice cream' into 'cake OR ice AND cream'
text = pattern_1.sub(r'\1 AND \2', text)

# make 'cake OR ice AND cream' into 'cake OR (ice AND cream)'
text = pattern_2.sub(r'(\1)', text)
print(text)

Input:

cake OR ice cream

Output:

cake OR (ice AND cream)

Answer 2

You can use a pattern to match TO_PARENTHESIS surrounded by a word, and then in the callback of re.sub place the full match between parenthesis and replace TO_PARENTHESIS by AND

\w+(?:\s+TO_PARENTHESIS\s+\w+)+

The pattern matches:

• \\w+ Match 1+ word characters
• (?: Non capture group
  • \\s+TO_PARENTHESIS Match whitespace chars and TO_PARENTHESIS
  • \\s+\\w+ Match whitespace chars and 1+ word chars
• )+ Close non capture group and repeat 1 or more times for multiple matches

Regex demo

import regex as re

text = 'cake OR ice cream please'
text = re.sub(r'(?<!OR|AND|NOT|\(|\)|")\s+(?!OR|AND|NOT|\(|\)|")', r' TO_PARENTHESIS ', text)
text = re.sub(
    r"\w+\s+TO_PARENTHESIS\s+\w+",
    lambda x: "(" + x.group().replace("TO_PARENTHESIS", "AND") + ")",
    text
)
print(text)

Output

cake OR (ice AND cream)

See a Python demo .

---

If the input is

cake OR ice cream please

The output will be

cake OR (ice AND cream AND please)

If you want to do a single replacement only instead of multiple, you can shorten the pattern to:

\w+\s+TO_PARENTHESIS \w+