以交替方式组合两个列表的 Pythonic 方式？
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Pythonic way to combine two lists in an alternating fashion?
提问by davidchambers
I have two lists, the first of which is guaranteed to contain exactly one more item than the second. I would like to know the most Pythonic way to create a new list whose even-index values come from the first list and whose odd-index values come from the second list.
我有两个列表，其中第一个保证包含比第二个多一个项目。我想知道创建一个新列表的最 Pythonic 方法，该列表的偶数索引值来自第一个列表，其奇数索引值来自第二个列表。
# example inputs
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
# desired output
['f', 'hello', 'o', 'world', 'o']
This works, but isn't pretty:
这有效，但并不漂亮：
list3 = []
while True:
try:
list3.append(list1.pop(0))
list3.append(list2.pop(0))
except IndexError:
break
How else can this be achieved? What's the most Pythonic approach?
这还能如何实现？最 Pythonic 的方法是什么？
采纳答案by Duncan
Here's one way to do it by slicing:
这是通过切片来实现的一种方法：
>>> list1 = ['f', 'o', 'o']
>>> list2 = ['hello', 'world']
>>> result = [None]*(len(list1)+len(list2))
>>> result[::2] = list1
>>> result[1::2] = list2
>>> result
['f', 'hello', 'o', 'world', 'o']
回答by David Z
There's a recipe for this in the itertoolsdocumentation:
itertools文档中有一个配方：
from itertools import cycle, islice
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
pending = len(iterables)
nexts = cycle(iter(it).next for it in iterables)
while pending:
try:
for next in nexts:
yield next()
except StopIteration:
pending -= 1
nexts = cycle(islice(nexts, pending))
回答by Mark Byers
This should do what you want:
这应该做你想做的：
>>> iters = [iter(list1), iter(list2)]
>>> print list(it.next() for it in itertools.cycle(iters))
['f', 'hello', 'o', 'world', 'o']
回答by wheaties
I'd do the simple:
我会做简单的：
chain.from_iterable( izip( list1, list2 ) )
It'll come up with an iterator without creating any additional storage needs.
它将提供一个迭代器，而不会产生任何额外的存储需求。
回答by Tom Anderson
I'm too old to be down with list comprehensions, so:
我太老了，不能理解列表，所以：
import operator
list3 = reduce(operator.add, zip(list1, list2))
回答by Jay
Here's a one liner that does it:
这是一个这样做的班轮：
list3 = [ item for pair in zip(list1, list2 + [0]) for item in pair][:-1]
list3 = [ item for pair in zip(list1, list2 + [0]) for item in pair][:-1]
回答by chernevik
Here's a one liner using list comprehensions, w/o other libraries:
这是一个使用列表推导式的单行代码，没有其他库：
list3 = [sub[i] for i in range(len(list2)) for sub in [list1, list2]] + [list1[-1]]
Here is another approach, if you allow alteration of your initial list1 by side effect:
这是另一种方法，如果您允许通过副作用更改初始 list1：
[list1.insert((i+1)*2-1, list2[i]) for i in range(len(list2))]
回答by Carlos Valiente
My take:
我的看法：
a = "hlowrd"
b = "el ol"
def func(xs, ys):
ys = iter(ys)
for x in xs:
yield x
yield ys.next()
print [x for x in func(a, b)]
回答by jwp
Stops on the shortest:
在最短的时间停止：
def interlace(*iters, next = next) -> collections.Iterable:
"""
interlace(i1, i2, ..., in) -> (
i1-0, i2-0, ..., in-0,
i1-1, i2-1, ..., in-1,
.
.
.
i1-n, i2-n, ..., in-n,
)
"""
return map(next, cycle([iter(x) for x in iters]))
Sure, resolving the next/__next__ method may be faster.
当然，解析 next/__next__ 方法可能会更快。
回答by killown
def combine(list1, list2):
lst = []
len1 = len(list1)
len2 = len(list2)
for index in range( max(len1, len2) ):
if index+1 <= len1:
lst += [list1[index]]
if index+1 <= len2:
lst += [list2[index]]
return lst
