根據python中的特殊字符將動態列表拆分爲子列表

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我在python中有一個基本問題，那就是我試圖長時間找到解決方案，但是我無法獲得正確的輸出。根據python中的特殊字符將動態列表拆分爲子列表

textvalues=[['1 of 2 DOCUMENTS', 'The New York Times', 'March 17, 2016 Thursday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section A; Column 0; Classified; Pg. 19', 'LENGTH: 176 words', 'LOAD-DATE: March 17, 2016', 'Copyright 2016 The New York Times Company', '', '2 of 2 DOCUMENTS', 'The New York Times', 'March 16, 2016 Wednesday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section B; Column 0; Classified; Pg. 16', 'LENGTH: 176 words', 'LOAD-DATE: March 16, 2016', 'Copyright 2016 The New York Times Company']]

這裏我需要根據「特殊字符」將上面的列表拆分成子列表。上面的列表是樣本列表，主列表是動態的，列表的長度可能不同。在任何情況下，列表都需要用「'字符分隔。

解決方案，我曾嘗試：

MainText = str(textvalues) 
split_index = MainText.index('',) 
l2 = MainText[:split_index] 
print(l2)

預期的解決方案：

[['1 of 2 DOCUMENTS', 'The New York Times', 'March 17, 2016 Thursday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section A; Column 0; Classified; Pg. 19', 'LENGTH: 176 words', 'LOAD-DATE: March 17, 2016', 'Copyright 2016 The New York Times Company'] ,['2 of 2 DOCUMENTS', 'The New York Times', 'March 16, 2016 Wednesday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section B; Column 0; Classified; Pg. 16', 'LENGTH: 176 words', 'LOAD-DATE: March 16, 2016', 'Copyright 2016 The New York Times Company']]

請幫我解決這個問題。由於

來源

2016-12-07 Mho

檢查右腿的解決方案。它適用於一些修改。在他的回答的評論中看到我的代碼。 – MYGz

檢查我的解決方案，如果它適合你。 – MYGz

import itertools 

textvalues=[['1 of 2 DOCUMENTS', 'The New York Times', 'March 17, 2016 Thursday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section A; Column 0; Classified; Pg. 19', 'LENGTH: 176 words', 'LOAD-DATE: March 17, 2016', 'Copyright 2016 The New York Times Company', '', '2 of 2 DOCUMENTS', 'The New York Times', 'March 16, 2016 Wednesday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section B; Column 0; Classified; Pg. 16', 'LENGTH: 176 words', 'LOAD-DATE: March 16, 2016', 'Copyright 2016 The New York Times Company']] 
groups = [] 
for a,b in itertools.groupby(textvalues[0], lambda x: x is not ''): 
    if a: 
     groups.append(list(b)) 
print groups

輸出：

[['1 of 2 DOCUMENTS', 'The New York Times', 'March 17, 2016 Thursday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section A; Column 0; Classified; Pg. 19', 'LENGTH: 176 words', 'LOAD-DATE: March 17, 2016', 'Copyright 2016 The New York Times Company'], ['2 of 2 DOCUMENTS', 'The New York Times', 'March 16, 2016 Wednesday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section B; Column 0; Classified; Pg. 16', 'LENGTH: 176 words', 'LOAD-DATE: March 16, 2016', 'Copyright 2016 The New York Times Company']]

來源

2016-12-07 04:03:09 MYGz

好的解決方案。非常棘手。感謝分享它。 –

基本上，你可以遍歷的內容，存儲在緩衝區中的子串，並轉儲緩衝區主列表跨越''分離器何時到來：

result = list() 
line = list() 
for element in textvalues[0]: 
    if element != '': 
     line.append(element) 
    else: 
     result.append(line) 
     line = list()

來源

2016-12-07 04:43:53

修復您的解決方案。檢查並編輯你的答案。 'textvalues = [['asd'，''，'asd d'，''，'c as d'，''，'asd f'，''，'lskd']] result = [] line = [] 爲元件在textvalues [0]：如果元素= ''： line.append（元件）否則： result.append（線）線= [] 否則： result.append（線）打印結果' – MYGz

上述代碼的輸出：'[['asd']，['asd d']，['c as d']，['asd f']，['lskd']]' – MYGz

它引發錯誤，因爲多個其他人在那裏。 – Mho

textvalues=[['1 of 2 DOCUMENTS', 'The New York Times', 'March 17, 2016 Thursday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section A; Column 0; Classified; Pg. 19', 'LENGTH: 176 words', 'LOAD-DATE: March 17, 2016', 'Copyright 2016 The New York Times Company', '', '2 of 2 DOCUMENTS', 'The New York Times', 'March 16, 2016 Wednesday\xa0\xa0Late Edition - Final', 'Paid Notice: Deaths THORNTON, ROBERT', 'SECTION: Section B; Column 0; Classified; Pg. 16', 'LENGTH: 176 words', 'LOAD-DATE: March 16, 2016', 'Copyright 2016 The New York Times Company']] 

textvalues2 = [] 

for i in ','.join(i for i in textvalues[0]).split(',,') : 
    textvalues2.append(i.split(','))

來源

2016-12-07 06:33:02 jwdasdk

根據python中的特殊字符將動態列表拆分爲子列表

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