在 Python 列表中移动值但保留顺序
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Moving values but preserving order in a Python list
提问by sloth
I have a list
我有一个清单
a=[1,2,3,4,5]
and want to 'move' its values so it changes into
并希望“移动”它的值，使其变为
a=[2,3,4,5,1]
and the next step
和下一步
a=[3,4,5,1,2]
Is there a built-in function in Python to do that?
Python 中是否有内置函数可以做到这一点？
Or is there a shorter or nicer way than
或者有比这更短或更好的方法
b=[a[-1]]; b.extend(a[:-1]); a=b
回答by Miles
>>> a = [1,2,3,4,5]
>>> a.append(a.pop(0))
>>> a
[2, 3, 4, 5, 1]
This is expensive, though, as it has to shift the contents of the entire list, which is O(n). A better choice may be to use collections.dequeif it is available in your version of Python, which allow objects to be inserted and removed from either end in approximately O(1) time:
但是，这很昂贵，因为它必须移动整个列表的内容，即 O(n)。collections.deque如果它在您的 Python 版本中可用，则更好的选择可能是使用它，它允许在大约 O(1) 时间内从任一端插入和删除对象：
>>> a = collections.deque([1,2,3,4,5])
>>> a
deque([1, 2, 3, 4, 5])
>>> a.rotate(-1)
>>> a
deque([2, 3, 4, 5, 1])
Note also that both these solutions involve changing the original sequence object, whereas yours creates a new list and assigns it to a. So if we did:
另请注意，这两种解决方案都涉及更改原始序列对象，而您的解决方案会创建一个新列表并将其分配给a. 所以如果我们这样做：
>>> c = a
>>> # rotate a
With your method, cwould continue to refer to the original, unrotated list, and with my methods, it will refer to the updated, rotatedlist/deque.
使用您的方法，c将继续引用原始的、未旋转的列表，而使用我的方法，它将引用更新的、旋转的列表/双端队列。
