問題定義將每一行分成句子。假設以下字符分隔句子：句點 ('.')、問號 ('?') 和感嘆號 ('!')。這些定界符也應該從返回的句子中省略。刪除每個句子中的任何前導或尾隨空格。如果在上述之后，一個句子是空白的（空字符串，''），則應該省略該句子。返回句子列表。句子的順序必須與它們在文件中出現的順序相同。這是我當前的代碼import redef get_sentences(doc):    assert isinstance(doc, list)    result = []    for line in doc:        result.extend(            [sentence.strip() for sentence in re.split(r'\.|\?|\!', line) if sentence]        )    return result# Demo:get_sentences(demo_input)輸入demo_input = ["  This is a phrase; this, too, is a phrase. But this is another sentence.",                  "Hark!",                  "    ",                  "Come what may    <-- save those spaces, but not these -->    ",                  "What did you say?Split into 3 (even without a space)? Okie dokie."]期望的輸出["This is a phrase; this, too, is a phrase", "But this is another sentence", "Hark", "Come what may    <-- save those spaces, but not these -->", "What did you say", "Split into 3 (even without a space)", "Okie dokie"]但是，我的代碼產生了這個：['This is a phrase; this, too, is a phrase', 'But this is another sentence', 'Hark', '', 'Come what may    <-- save those spaces, but not these -->', 'What did you say', 'Split into 3 (even without a space)', 'Okie dokie']問題：為什么''即使我的代碼忽略了它，我也會在其中得到那個空句子？我可以使用以下代碼解決問題，但我將不得不再次瀏覽列表，我不想這樣做。我想在同一個過程中做到這一點。import redef get_sentences(doc):    assert isinstance(doc, list)    result = []    for line in doc:        result.extend([sentence.strip() for sentence in re.split(r'\.|\?|\!', line)])        result = [s for s in result if s]    return result# Demo:get_sentences(demo_input)