CODEWITHSUNDEEP
pythonregexparsing
Crusader
Crusader

Reputation: 333

Create list of lists based on a condition

I have a string like below:

result = """The following table provides the details.
acquired, by major class:
(US$ in millions)    Customer relationships       15year       $265
There is another line without space here.
Another table starts here:
(USS in millions)       2018       2017
Income (loss) from continuing operations       $298       $129"""

I have to take all the sentences that contain more than 3 spaces and put them in a list of lists. Below is something I have tried so far:

lines = result.splitlines()
table_list = []
for i in range(len(lines)):
    if re.search(r'   {3,}', lines[i]):
        table_list.append(lines[i])

Resultant output of above code:

['(US$ in millions)       Customer relationships      15year      $265','(USS in millions)     2018      2017','Income (loss) from continuing operations       $298      $129']

Expected Output:

[['(US$ in millions)       Customer relationships      15year      $265'],['(USS in millions)       2018       2017','Income (loss) from continuing operations       $298       $129']]

Further explanation of output condition: Expected output should be a list of lists. While iterating through each line, if there are consecutive sentences that contain 3 or more spaces between 2 words, all of these lines should be part of same list within the main list. If a line does not contain 3 or more spaces between 2 words, this breaks the chain. If there is another line that contains 3 or more spaces between 2 words then this line becomes part of a new list inside the main list.

Upvotes: 0

Views: 51

Answers (1)

Chris
Chris

Reputation: 29742

Use itertools.groupby with re.findall:

from itertools import groupby

def has_spaces(str_):
    return bool(re.findall("\s{3,}", str_))

[list(g) for k, g in groupby(result.splitlines(), key=has_spaces) if k]

Output:

[['(US$ in millions)    Customer relationships       15year       $265'],
 ['(USS in millions)       2018       2017',
  'Income (loss) from continuing operations       $298       $129']]

Upvotes: 1

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