正则表达式负向后看python

Question

I'm trying to split a string by commas that are not inside brackets (ie the string contains items that are separated by commas, but it also contains commas within brackets that I don't want to separate on). Like so:

A='[1, "A"], [2, "B"], [3, "C"], [4, "D"], [5, "E"], [6, "F"], [7, "G"], [8, "H"], [9, "I"], [10, "J"], [100, "JJ"]'

Which should result in:

['[1, "A"]', ' [2, "B"]', ' [3, "C"]', ' [4, "D"]', ' [5, "E"]', ' [6, "F"]', ' [7, "G"]', ' [8, "H"]', ' [9, "I"]', ' [10, "J"]', '[100, "JJ"]']

I tried using negative lookbehind like this:

B=re.split(r'(?<![[][\d]),',A)

However, this does not work when the number within the brackets goes above one digit such as in the case of [10, "J"]. Any help would be greatly appreciated!

Answer 1

This looks like "split on any comma that is preceded by a ] " could work. For good measure I added \\s* to eat up the spaces before the next item.

import re

A = '[1, "A"], [2, "B"], [3, "C"], [4, "D"], [5, "E"], [6, "F"], [7, "G"], [8, "H"], [9, "I"], [10, "J"], [100, "JJ"]'

re.split(r"(?<=]),\s*", A)

gives

['[1, "A"]', '[2, "B"]', '[3, "C"]', '[4, "D"]', '[5, "E"]', '[6, "F"]', '[7, "G"]', '[8, "H"]', '[9, "I"]', '[10, "J"]', '[100, "JJ"]']

Answer 2

You can try this:

A='[1, "A"], [2, "B"], [3, "C"], [4, "D"], [5, "E"], [6, "F"], [7, "G"], [8, "H"], [9, "I"], [10, "J"], [100, "JJ"]'
import re
data = re.split('(?<=\]),\s', A)

Output:

['[1, "A"]', '[2, "B"]', '[3, "C"]', '[4, "D"]', '[5, "E"]', '[6, "F"]', '[7, "G"]', '[8, "H"]', '[9, "I"]', '[10, "J"]', '[100, "JJ"]']

Answer 3

If using split is not a requirement, findall can also be used in with a very simple expression,

In [27]: re.findall(r'\[.+?\]', A)
Out[27]:
['[1, "A"]', '[2, "B"]', '[3, "C"]', '[4, "D"]', '[5, "E"]', '[6, "F"]', '[7, "G"]', '[8, "H"]', '[9, "I"]', '[10, "J"]', '[100, "JJ"]']

Answer 4

try this regex a get each item by group 1:

(\\[\\d+,\\s*\\"\\w+\\"\\])

You can see the result in this link:

https://regex101.com/r/K5XV6F/1

Answer 5

Using the newer regex module you can use

\[[^][]*\](*SKIP)(*FAIL) # discard anything in square brackets
|                        # or
,\s*                     # match , and whitespaces, eventually

---

In Python this looks like

 import regex as re A='[1, "A"], [2, "B"], [3, "C"], [4, "D"], [5, "E"], [6, "F"], [7, "G"], [8, "H"], [9, "I"], [10, "J"], [100, "JJ"]' rx = re.compile(r'\\[[^][]*\\](*SKIP)(*FAIL)|,\\s*') print(rx.split(A)) # ['[1, "A"]', '[2, "B"]', '[3, "C"]', '[4, "D"]', '[5, "E"]', '[6, "F"]', '[7, "G"]', '[8, "H"]', '[9, "I"]', '[10, "J"]', '[100, "JJ"]'] 

See a demo on regex101.com .