将列表拆分为第一个和休息的 Pythonic 方法？
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Pythonic way to split a list into first and rest?
提问by Shawn
I think in Python 3 I'll be able to do:
我认为在 Python 3 中我将能够做到：
first, *rest = l
which is exactly what I want, but I'm using 2.6. For now I'm doing:
这正是我想要的，但我使用的是 2.6。现在我正在做：
first = l[0]
rest = l[1:]
This is fine, but I was just wondering if there's something more elegant.
这很好，但我只是想知道是否有更优雅的东西。
采纳答案by Shawn
first, rest = l[0], l[1:]
Basically the same, except that it's a oneliner. Tuple assigment rocks.
基本上是一样的，只是它是一个单线。元组分配岩石。
This is a bit longer and less obvious, but generalized for all iterables (instead of being restricted to sliceables):
这有点长且不太明显，但适用于所有可迭代对象（而不是仅限于切片）：
i = iter(l)
first = next(i) # i.next() in older versions
rest = list(i)
回答by Olivier '?lbaum' Scherler
You can do
你可以做
first = l.pop(0)
and then lwill be the rest. It modifies your original list, though, so maybe it's not what you want.
然后l就是剩下的了。但是，它会修改您的原始列表，因此它可能不是您想要的。
回答by FGH
I would suggest:
我会建议：
first, remainder = l.split(None, maxsplit=1)
回答by kriss
Yet another one, working with python 2.7. Just use an intermediate function. Logical as the new behavior mimics what happened for functions parameters passing.
另一个，使用 python 2.7。只需使用一个中间函数。合乎逻辑，因为新行为模仿了函数参数传递时发生的情况。
li = [1, 2, 3]
first, rest = (lambda x, *y: (x, y))(*li)
