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pythonregex
Tony Montana
Tony Montana

Reputation: 1018

Using regex to remove the unnecessary whitespace to get the expected output

I have a few unstructured data like this

test1     21;
 test2  22;
test3    [ 23 ];

and I want to remove the unnecessary whitespace and convert it into the list of two-item per row and the expected output should look like this

['test1', '21']
['test2', '22']
['test3', ['23']]

Now, I am using this regex sub method to remove the unnecessary whitespace

re.sub(r"\s+", " ", z.rstrip('\n').lstrip(' ').rstrip(';')).split(' ')

Now, the problem is that it is able to replace the unnecessary whitespace into single whitespace, which is fine. But the problem I am facing in the third example, where after and before the open and close bracket respectively, it has whitespace and that I what to remove. But using the above regex I am not able to.

This is the output currently I am getting

['test1', '21']
['test2', '22']
['test3', '[', '23', ']']

You may check the example here on pythontutor.

Upvotes: 3

Views: 128

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626893

You can use

import re
 
x = "test1     21"
y = "     test2  22"
z = "    test3    [ 23 ]"
 
for a in [x, y, z]:
    print(re.sub(r"(?<![^[\s])\s+|\s+(?=])", "", a.rstrip('\n').lstrip(' ').rstrip(';')).split(' '))

See the Python demo. Output:

['test1', '21']
['test2', '22']
['test3', '[23]']

Details:

  • (?<![^[\s])\s+ - one or more whitespaces that are preceded with a [ char, whitespace or start of string
  • | - or
  • \s+(?=]) - one or more whitespaces that are followed with a ] char.

Upvotes: 1

anubhava
anubhava

Reputation: 785216

You may use this regex with 2 capture groups:

(\w+)\s+(\[[^]]+\]|\w+);

RegEx Demo

RegEx Details:

  • (\w+): Match 1+ word characters in first capture group
  • \s+: Match 1+ whitespaces
  • (\[[^]]+\]|\w+): Match a [...] string or a word in second capture group
  • ;: Match a ;

Code:

>>> import re
>>> data = '''
... test1     21;
...  test2  22;
... test3    [ 23 ];
... '''
>>> res = []
>>>
>>> for i in re.findall(r'(\w+)\s+(\[[^]]+\]|\w+);', data):
...     res.append([ i[0], eval(re.sub(r'^(\[)\s*|\s*(\])$', r'\1"\2', i[1])) if i[1].startswith('[') else i[1] ])
...
>>> print (res)
[['test1', '21'], ['test2', '22'], ['test3', ['23']]]

Upvotes: 2

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