Python:如何基于特定元素拆分列表

Question

如果我们在Python中有以下列表

sentence = ["I", "am", "good", ".", "I", "like", "you", ".", "we", "are", "not", "friends", "."]

如何将其拆分以获得包含以句点结束的元素的列表？所以我想在我的新列表中获得以下元素:

["I","am","good","."]
["I","like","you","."]
["we","are","not","friends","."]

到目前为止我的尝试:

cleaned_sentence = []
a = 0
while a < len(sentence):
    current_word = sentence[a]
    if current_word == "." and len(cleaned_sentence) == 0:
        cleaned_sentence.append(sentence[0:sentence.index(".")+1])
        a += 1
    elif current_word == "." and len(cleaned_sentence) > 0:
        sub_list = sentence[sentence.index(".")+1:-1]
        sub_list.append(sentence[-1])
        cleaned_sentence.append(sub_list[0:sentence.index(".")+1])
        a += 1
    else:
        a += 1

for each in cleaned_sentence:
    print(each)

运行此sentence生成

['I', 'am', 'good', '.']
['I', 'like', 'you', '.']
['I', 'like', 'you', '.']

Answer 1

你可以使用itertools.groupby:

from itertools import groupby
i = (list(g) for _, g in groupby(sentence, key='.'.__ne__))
print([a + b for a, b in zip(i, i)])

这输出:

[['I', 'am', 'good', '.'], ['I', 'like', 'you', '.'], ['we', 'are', 'not', 'friends', '.']]

如果您的列表并不总是以结束,'.'那么您可以使用itertools.zip_longest:

sentence = ["I", "am", "good", ".", "I", "like", "you", ".", "we", "are", "not", "friends"]
i = (list(g) for _, g in groupby(sentence, key='.'.__ne__))
print([a + b for a, b in zip_longest(i, i, fillvalue=[])])

这输出:

[['I', 'am', 'good', '.'], ['I', 'like', 'you', '.'], ['we', 'are', 'not', 'friends']]