Question

这就是我所拥有的：

item_dictionary = {'10350': 'Third-Age Full', '560':'Death Rune'}

filler = []
for i in range(0,365):
    filler.append(None)

filler1=[]
for i in range(0, 2):
    filler1.append(None)

filler2=[]
for i in range(0, 1):
    filler2.append(None)

names = item_dictionary.values()
names_filled = list(itertools.chain.from_iterable(zip(names, filler1 * len(names))))
#Returns ['Death Rune', None, 'Third-Age Full', None]

和

names = item_dictionary.values()
names1 = list(itertools.chain.from_iterable(zip(names, filler2 * len(names))))
names_filled = list(itertools.chain.from_iterable(zip(names1, filler1 * len(names))))
#Returns ['Death Rune', None, None, None, 'Third-Age Full', None, None, None]

我正试图获得['Death Rune', None, None, 'Third-Age Full', None, None]，但我似乎无法，有人可以帮助我吗？感谢。

Answer 1

怎么样：

In [171]: item_dictionary = {'10350': 'Third-Age Full', '560':'Death Rune'}

In [172]: names = item_dictionary.values()

In [173]: [n for name in names for n in [name, None, None]]
Out[173]: ['Death Rune', None, None, 'Third-Age Full', None, None]

或者如果您想轻松将None的数量更改为其他值，

In [174]: [n for name in names for n in [name] + [None]*2]
Out[174]: ['Death Rune', None, None, 'Third-Age Full', None, None]

Answer 2

首先，您可以更轻松地创建这些填充物：

filler = [None] * 365
filler1 = [None] * 2
filler2 = [None]

要真正回答你的问题，你可以压缩两个以上的迭代，所以你可以将第三个填充物压缩到它：

>>> list(itertools.chain.from_iterable(zip(names, [None] * len(names), [None] * len(names))))
['Third-Age Full', None, None, 'Death Rune', None, None]

然后你也可以使用itertools.zip_longest根本不需要你扩展那些填充迭代，但只是自动扩展它们（zip_longest将使用None作为其每个默认值为fillvalue）：

>>> list(itertools.chain.from_iterable(itertools.zip_longest(names, [], [])))
['Third-Age Full', None, None, 'Death Rune', None, None]

或者，正如unutbu建议的那样，使用列表理解。

Answer 3

这不是你想做的吗？

names_filled = list(itertools.chain.from_iterable([[n,None,None] for n in names])

我可以获得1或3'无'值，但不能获得2

3 个答案: