Question

我在python中有一个基本问题，我试图找到解决方案很长一段时间，但我无法得到正确的输出。

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']]

这里我需要根据''特殊字符将上面的列表拆分为子列表。以上列表是样本列表，主列表是动态的，其中列表的长度可能不同。在任何情况下，列表都需要用''字符分隔。

我尝试过的解决方案：

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

预期解决方案：

[['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']]

请帮我解决这个问题。感谢

Answer 1

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

输出：

[['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']]

Answer 2

基本上，您可以遍历内容，将子字符串存储在缓冲区中，并在遇到''分隔符时将缓冲区转储到主列表：

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

Answer 3

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(',') )

根据python中的特殊字符将动态列表拆分为子列表

3 个答案: