根據條件創建列表列表

Question

我有一個如下所示的字符串：

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"""

我必須把所有包含超過 3 個空格的句子放在一個列表列表中。 以下是我到目前為止嘗試過的東西：

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

以上代碼的結果 output：

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

預計 Output：

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

output 條件的進一步說明： Expected output should be a list of lists 。 在遍歷每一行時，如果有連續的句子在 2 個單詞之間包含 3 個或更多空格，則所有這些行都應該是主列表中同一列表的一部分。 如果一行在 2 個單詞之間不包含 3 個或更多空格，則鏈會中斷。 如果有另一行在 2 個單詞之間包含 3 個或更多空格，則該行將成為主列表中新列表的一部分。

Answer 1

將itertools.groupby與re.findall一起使用：

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']]