Question

我正在尝试从网页中删除此列表中的“title”元素：

x = 
[[(u'title', u'Goals for')], [(u'title', u'Goals against')], [(u'title', u'Penalty goal')], [(u'title', u'Goals for average')], [(u'title', u'Matches Played')], [(u'title', u'Shots on goal')], [(u'title', u'Shots Wide')], [(u'title', u'Free Kicks Received')], [(u'title', u'Offsides')], [(u'title', u'Corner kicks')], [(u'title', u'Wins')], [(u'title', u'Draws')], [(u'title', u'Losses')]]

我希望我的决赛能够

result = ['Goals for', 'Goals against','Penalty goal','Goals for average',....]

但我可以做y = x[1][0][1] =＆gt; '目标' 我不能不x[i][0][1]因为它是我的for循环语句中的索引我得到了错误

TypeError：列表索引必须是整数，而不是元组

我怎么能解决这个问题？

Answer 1

我会使用列表理解：

>>> new = [sublist[0][1] for sublist in x]
>>> pprint.pprint(new)
[u'Goals for',
 u'Goals against',
 u'Penalty goal',
 u'Goals for average',
 u'Matches Played',
 u'Shots on goal',
 u'Shots Wide',
 u'Free Kicks Received',
 u'Offsides',
 u'Corner kicks',
 u'Wins',
 u'Draws',
 u'Losses']

不确定pandas连接是什么。如果您尝试从MultiIndex中提取列，则有更简单的方法。

Answer 2

你可以使用列表理解（通常更常见，因为它清晰，简洁并被认为是Pythonic）：

x = [[(u'title', u'Goals for')], [(u'title', u'Goals against')], [(u'title', u'Penalty goal')], [(u'title', u'Goals for average')], [(u'title', u'Matches Played')], [(u'title', u'Shots on goal')], [(u'title', u'Shots Wide')], [(u'title', u'Free Kicks Received')], [(u'title', u'Offsides')], [(u'title', u'Corner kicks')], [(u'title', u'Wins')], [(u'title', u'Draws')], [(u'title', u'Losses')]]
x = [i[0][1:] for i in x]

或者您可以使用for上的x循环：

for i in range(len(x)):
    x[i] = x[i][0][1:]

正如原始答案之后所指出的那样，我使用Python的del语句（例如del x[0][0][0]）的其他原始建议也不起作用，因为tuple不支持项目删除

Answer 3

试一试：

x = [[('title', 'Goals for')], [('title', 'Goals against')], [('title', 'Penalty goal')], [('title', 'Goals for average')], [('title', 'Matches Played')], [('title', 'Shots on goal')], [('title', 'Shots Wide')], [('title', 'Free Kicks Received')], [('title', 'Offsides')], [('title', 'Corner kicks')], [('title', 'Wins')], [('title', 'Draws')], [('title', 'Losses')]]
print([element[0][1] for element in x ])

Answer 4

其他解决方案：

>>> map(lambda a: a[0][1], x)
... [u'Goals for', u'Goals against', u'Penalty goal', u'Goals for average', u'Matches Played', u'Shots on goal', u'Shots Wide', u'Free Kicks Received', u'Offsides', u'Corner kicks', u'Wins', u'Draws', u'Losses']
>>>

取消列出python列表并删除元素

4 个答案: