Special regex for String in python

Question

I have a string like below.

s = ({[test1, test2 ; test3 (New) ]})

Now I have a regex which will remove brackets and convert it into the list. Even if there are separated with a;b,c like. REGEX:

output = [i for i in re.split(r'\s*[(){}<>\[\],;\'"]\s*', s) if i]

But this regex is removing brackets from items of the list as well. ((New) in my case)

How to apply this regex for beginnig and end of the string. I know it can be done using ^ but not sure how?

Expected Output

['test1', 'test2', 'test3 (New)' ]

Output coming from above regex

['test1', 'test2', 'test3', 'New']

Any help?

Is adding the brackets to the list element an option ? Is the `(New)` element always the last in the list ? — David, May 22 '18 at 09:10
No, I mean there may be `(New)` something like this in any element of my string. I want to remove brackets only from starting and end of string — Jayesh Dhandha, May 22 '18 at 09:12
@JayeshDhandha, what should be the result for this string `-[({[test1, test2(3) ; test3 (New) ]})]-`? — RomanPerekhrest, May 22 '18 at 09:32

SpghttCd · Accepted Answer · 2018-05-22T11:16:42.397

1

s = '({[test1, test2 ; test3 (New) ]})'

Based on your comment below I assume that the number of opening brackets of the whole string is equal to the number of closing brackets.

So removing the outer brackets first needs to know their number:

m = re.match('[({[]*', s)
n_brckt = m.span()[1] - m.span()[0]

Then remove the outer brackets ( - dependent on if there were found any...):

if n_brckt > 0:
    s = s[n_brckt:-n_brckt]
s = s.strip()

In: s
Out: 'test1, test2 ; test3 (New)'

Then you can split at all occurences of commas or colons optionally followed by a space:

In: re.split('[,;]+ *', s)
Out: ['test1', 'test2', 'test3 (New)']

edited May 22 '18 at 11:16

answered May 22 '18 at 09:13

SpghttCd

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I can't use `strip` as there may be different number of brackets in the starting and ending of string. I can't use `string with 3 count` – Jayesh Dhandha May 22 '18 at 09:18
So how _do_ the beginning and the end of the strings look like? This should be made clear, otherwise it's just a guessing game... – SpghttCd May 22 '18 at 09:20
Its not fix. Sometimes it has `[{( ` and sometimes `[{` only. And same thing at the end of string. – Jayesh Dhandha May 22 '18 at 09:23
You have missed 1 thing. If my string doesn't contain any brackets then your output will be `['']` Which is not right. – Jayesh Dhandha May 22 '18 at 09:48
Do you want to tell me, that the string might even have no outer brackets at all...? – SpghttCd May 22 '18 at 10:55
Yes. We have to see all the corner case :) – Jayesh Dhandha May 22 '18 at 11:03

score 1 · Answer 2 · answered May 22 '18 at 10:05

1

Using re.search

import re
s = "({[test1, test2 ; test3 (New) ]})"
m = re.search("\[(.*?)\]", s)
if m:
    #print(m.group(1).replace(";", ",").split(",")) 
    print([i.strip() for i in m.group(1).replace(";", ",").split(",")])

Output:

['test1', 'test2', 'test3 (New)']

answered May 22 '18 at 10:05

Rakesh

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Good Answer. Thanks! – Jayesh Dhandha May 22 '18 at 10:09
Check output for `(({{[[test1, test2 ; test3 (France)]]}}))` – Jayesh Dhandha May 22 '18 at 10:11
1

For a deep nested structure, regex will not work and can cause unwanted issues. It is best to find a parser according to your need. Ex: https://stackoverflow.com/questions/5454322/python-how-to-match-nested-parentheses-with-regex – Rakesh May 22 '18 at 10:13
Please note that `if m:` is not a valid test for success of `re.match`. You _could_ use `if m.end():` but however, clearly writing `if m.end() > 0:` or sth similar is always clearer to read and understand immediately. – SpghttCd May 22 '18 at 11:27

Special regex for String in python

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