如何在 Python 中使用正则表达式匹配和替换多个字符串

Question

I am trying to replace some text in Python with regex.

My text looks like this:

WORKGROUP 1. John Doe ID123, Jane Smith ID456, Ohe Keedoke ID7890
Situation paragraph 1

WORKGROUP 2. John Smith ID321, Jane Doe ID654
Situation paragraph 2

What I am trying to do is put the names in double square brackets and remove the IDs so that it will end up looking like this.

WORKGROUP 1. [[John Doe]], [[Jane Smith]], [[Ohe Keedoke]]
Situation paragraph 1

WORKGROUP 2. [[John Smith]], [[Jane Doe]]
Situation paragraph 2

So far I have this.

re.sub(r"(WORKGROUP\s\d\.\s)",r"\1[[")
re.sub(r"(WORKGROUP\s\d\..+?)(?:\s\b\w+\b),(?:\s)(.+\n)",r"\1]], [[\2")
re.sub(r"(WORKGROUP\s\d\..+?)(?:\s\b\w+\b)(\n)",r"\1]]\2")

This works for groups with two people (WORKGROUP 2) but leaves all the IDs except the first and last persons' if there are more than two. So WORKGROUP 1 ends up looking like this.

WORKGROUP 1. [[John Doe]], [[Jane Smith ID456, Ohe Keedoke]]
Situation paragraph 1

Unfortunately, I can't do something like

re.sub(r"((\s\b\w+\b),(\s))+",r"\1]], [[\2")

because it will match inside the situation paragraphs.

My question is: is it possible to do multiple match/replacements in a string segment without doing it universally?

Answer 1

If you have the regex module installed:

(?<=\bWORKGROUP\s+\d+\.\s|,)\s*(.+?)\s*ID\d+\s*(?=,|$)

might work OK.

If not, you can simply do that in your terminal, by running:

$ pip install regex

or

$ pip3 install regex

Here, we're assuming that you might have other ID\d+ present in your text, otherwise, if you don't your problem would be much simple.

Test

import regex as re

regex = r"(?<=\bWORKGROUP\s+\d+\.\s|,)\s*(.+?)\s*ID\d+\s*(?=,|$)"

test_str = '''

WORKGROUP 1. John Doe ID123, Jane Smith ID456, Ohe Keedoke ID7890
Situation paragraph 1
WORKGROUP 2. John Smith ID321, Jane Doe ID654
Situation paragraph 2

WORKGROUP 11. Bob Doe ID123, Alice Doe ID123, John Doe ID123, Jane Smith ID456, Ohe Keedoke ID7890
Situation paragraph 1

WORKGROUP 21. John Smith ID321, Jane Doe ID654
Situation paragraph 2

'''


subst = "[[\\1]]"

print(re.sub(regex, subst, test_str, 0, re.MULTILINE))

Output

WORKGROUP 1. [[John Doe]],[[Jane Smith]],[[Ohe Keedoke]]
Situation paragraph 1
WORKGROUP 2. [[John Smith]],[[Jane Doe]]
Situation paragraph 2

WORKGROUP 11. [[Bob Doe]],[[Alice Doe]],[[John Doe]],[[Jane Smith]],[[Ohe Keedoke]]
Situation paragraph 1

WORKGROUP 21. [[John Smith]],[[Jane Doe]]
Situation paragraph 2

---

If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com . If you'd like, you can also watch in this link , how it would match against some sample inputs.

---

Answer 2

Code

import re

test = """
WORKGROUP 1. John Doe ID123, Jane Smith ID456, Ohe Keedoke ID7890
Situation paragraph 1

WORKGROUP 2. John Smith ID321, Jane Doe ID654
Situation paragraph 2
"""

test = re.sub(' ID[0-9]+, ', ']], [[', test)
test = re.sub('\. ', '. [[', test)
test = re.sub(' ID[0-9]+', ']]', test)
print(test)

Output

WORKGROUP 1. [[John Doe]], [[Jane Smith]], [[Ohe Keedoke]]
Situation paragraph 1

WORKGROUP 2. [[John Smith]], [[Jane Doe]]
Situation paragraph 2

Answer 3

You can nest the substitutions and make the first substitution find lines that start with WORKGROUP first, and then let the second substitution find and replace the common-separated tokens inside:

re.sub(
    r'^(WORKGROUP\s+\d+\.\s*)(.*)',
    lambda m: m.group(1) + re.sub(r'([^,\s][^,]*)\s+\S+(?=,|$)', r'[[\1]]', m.group(2)),
    text,
    flags=re.MULTILINE
)

so that given:

text = '''WORKGROUP 1. John Doe ID123, Jane Smith ID456, Ohe Keedoke ID7890
Situation paragraph 1

WORKGROUP 2. John Smith ID321, Jane Doe ID654
Situation paragraph 2'''

the expression returns:

WORKGROUP 1. [[John Doe]], [[Jane Smith]], [[Ohe Keedoke]]
Situation paragraph 1

WORKGROUP 2. [[John Smith]], [[Jane Doe]]
Situation paragraph 2

Demo: https://repl.it/@blhsing/BoldElderlyQuerylanguage