如何从右侧拆分单词/数字和符号模式处的字符串

Question

I am trying to split a string that may look like this :

A Fool (SEVEN000) (and His Money are S00n) Parted 

Into : A Fool (7000) (and His Money are and S00n) Parted using Python

The ) will always be present at the end of the string and will always be preceded by a word/number. I was thinking splitting it from the right using a [word/number]) pattern would work.

Edit :

As requested here are a few more examples

Right (Out of the) Gate 

Expected Output : Right (Out of the) Gate

Right (Out) (of the Gate at 12PM)

Expected Output : Right (Out of the Gate at 12PM)

Answer 1

You seem to be splitting your string from the last space present in your parenthesis. You can use this regex,

 (?=[^()]*\))(?=\S*\))

Demo

Check this Python code,

import re

s = 'A Fool (SEVEN000) (and His Money are S00n) Parted'
arr = re.split(r' (?=[^()]*\))(?=\S*\))', s)
print(arr)

Prints like you wanted,

['A Fool (SEVEN000) (and His Money are', 'S00n) Parted']

Answer 2

Here is one option using re.split with a positive lookahead. The pattern I use is:

\s+(?=\w+\)(?:\s|$))

This pattern says to split and consume any amount of whitespace, when what follows is one or more word characters which itself is followed by a closing parenthesis and whitespace or the end of the input.

input = "A Fool (SEVEN000) (and His Money are S00n) Parted"
parts = re.split(r'\s+(?=\w+\)(?:\s|$))', input)
print(parts)

['A Fool (SEVEN000) (and His Money are', 'S00n) Parted']

Answer 3

Use the following regular expression, and substring at the index:

\b[A-Za-z0-9]+\) [A-Za-z0-9]+$

(this assumes that after the closing bracket there is only a single word, you would need to give more information so I can update the answer if that's not true)

Answer 4

I would do it following way:

import re
text = 'A Fool (SEVEN000) (and His Money are S00n) Parted'
parted = re.findall(r'(.+)\s+(\S+\)[^\)]*$)',text)[0]
print(parted)

Output is following 2-tuple:

('A Fool (SEVEN000) (and His Money are', 'S00n) Parted')

To understand my regular expression it might be breaked into:

1st group: .+

seperator: \\s+

2nd group: \\S+\\)[^\\)]*$

First group match at least 1 characters not being newline \\n , seperator match at least 1 whitespace character (this mean not only space, but also carriage return \\r , tab \\t and so on), lastly but most importantly second group consist of at least one non-whitespace character followed by ) followed by 0 or more not- ) (ie any character which is not ) ) which spans to end of string as denoted by $ . If you want solely spaces instead whitespace characters then replace \\s with(space) and \\S with [^ ]

Answer 5

If the separator is only a whitespace, we can do it without regex. May be like this, using rfind() :

def splitter(a_string):
    idx1 = a_string.rfind(')')
    idx2 = a_string.rfind(' ', 0, idx1)
    idx3 = a_string.rfind('(', 0, idx1)
    if (idx2 > -1) and (idx3 < idx2):
         return (a_string[:idx2], a_string[idx2:])
    else:
         return None

input: splitter('Right (Out) (of the Gate at 12PM)')

output: ('Right (Out) (of the Gate at', ' 12PM)')

input: splitter('Right (Out)')

output: None