使用正则表达式拆分，但使用分隔符的第一个字符

Question

I have a regex like this: "[az|AZ|0-9]: " that will match one alphanumeric character, colon, and space. I wonder how to split the string but keeping the alphanumeric character in the first result of splitting. I cannot change the regex because there are some cases that the string will have special character before colon and space.

Example:

line = re.split("[a-z|A-Z|0-9]: ", "A: ") # Result: ['A', '']
line = re.split("[a-z|A-Z|0-9]: ", ":: )5: ") # Result: [':: )5', '']
line = re.split("[a-z|A-Z|0-9]: ", "Delicious :): I want to eat this again") # Result: ['Delicious :)', 'I want to eat this again']

Update: Actually, my problem is splitting from a review file. Suppose I have a file that every line has this pattern: [title]: [review] . I want to get the title and review, but some of the titles have a special character before a colon and space, and I don't want to match them. However, it seems that the character before a colon and space that I want to match apparently is an alphanumeric one.

Answer 1

Solution

First of all, as you show in your examples, you need to match characters other than a-zA-Z0-9 , so we should just use the . matcher, it will match every character.

So I think the expression you're looking for might be this one:

(.*?):(?!.*:) (.*)

You can use it like so:

import re

pattern = r"(.*?):(?!.*:) (.*)"
matcher = re.compile(pattern)

txt1 = "A: "
txt2 = ":: )5: "
txt3 = "Delicious :): I want to eat this again"

result1 = matcher.search(txt1).groups() # ('A', '')
result2 = matcher.search(txt2).groups() # (':: )5', '')
result3 = matcher.search(txt3).groups() # ('Delicious :)', 'I want to eat this again')

Explanation

We use capture groups (the parentheses) to get the different parts in the string into different groups, search then finds these groups and outputs them in the tuple.

The (?.:*:) part is called "Negative Lookahead", and we use it to make sure we start capturing from the last : we find.

Edit

BTW, if, as you mentioned, you have many lines each containing a review, you can use this snippet to get all of the reviews separated by title and body at once:

import re

pattern = r"(.*?):(?!.*:) (.*)\n?"
matcher = re.compile(pattern)

reviews = """ 
A: 
:: )5: 
Delicious :): I want to eat this again
"""

parsed_reviews = matcher.findall(reviews) # [('A', ''), (':: )5', ''), ('Delicious :)', 'I want to eat this again')]

Answer 2

You could split using a negative lookbehind with a single colon or use a character class [:)] where you can specify which characters should not occur directly to the left.

(?<!:):[ ]

In parts

• (?<::) Negative lookbehind, assert what is on the left is not a colon
• :[ ] Match a colon followed by a space (Added square brackets only for clarity)

Regex demo | Python demo

For example

import re
pattern = r"(?<!:): "
line = re.split(pattern, "A: ") # Result: ['A', '']
print(line)
line = re.split(pattern, ":: )5: ") # Result: [':: )5', '']
print(line)
line = re.split(pattern, "Delicious :): I want to eat this again") # Result: ['Delicious :)', 'I want to eat this again']
print(line)

Output

['A', '']
[':: )5', '']
['Delicious :)', 'I want to eat this again']