CODEWITHSUNDEEP
pythonregex
Estarossa
Estarossa

Reputation: 605

Python RegEx matching each bracket element

I have been trying to come up with a regex for the following string:

[1,null,"7. Mai 2017"],[2,"test","8. Mai 2018"],[3,"test","9. Mai 2019"]

I am trying to get as match output each bracket with its content as a single element like the following:

[1,null,"7. Mai 2017"]
[2,"test","8. Mai 2018"]
[3,"test","9. Mai 2019"]

My initial naive approach was something like this:

(\[[^d],.+\])+

However, the .+ rule is too general and ends up matching the whole line. Any hints?

Upvotes: 3

Views: 77

Answers (3)

dawg
dawg

Reputation: 104092

You might consider the wonderful module pyparsing to do this:

import pyparsing 

for match in pyparsing.originalTextFor(pyparsing.nestedExpr('[',']')).searchString(exp):
    print match[0]
[1,null,"7. Mai 2017"]
[2,"test","8. Mai 2018"]
[3,"test","9. Mai 2019"]

(Unless it is actually JSON -- use the JSON module if so...)

Upvotes: 1

alecxe
alecxe

Reputation: 474201

I am not sure about the data format you are trying to parse and where it is coming from, but it looks JSON-like. For this particular string, adding square brackets from the beginning and the end of the string makes it JSON loadable:

In [1]: data = '[1,null,"7. Mai 2017"],[2,"test","8. Mai 2018"],[3,"test","9. Mai 2019"]'

In [2]: import json

In [3]: json.loads("[" + data + "]")
Out[3]: 
[[1, None, u'7. Mai 2017'],
 [2, u'test', u'8. Mai 2018'],
 [3, u'test', u'9. Mai 2019']]

Note how null becomes Python's None.

Upvotes: 1

BernardoGO
BernardoGO

Reputation: 1856

The following code will output what you've requested using \[[^]]*].

import re
regex = r'\[[^]]*]'
line = '[1,null,"7. Mai 2017"],[2,"test","8. Mai 2018"],[3,"test","9. Mai 2019"]'
row = re.findall(regex, line)
print(row)

Output:

['[1,null,"7. Mai 2017"]', '[2,"test","8. Mai 2018"]', '[3,"test","9. Mai 2019"]']

Consider changing null to None as it matches python representation.

Upvotes: 1

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