list[:] = [...] 赋值时的冒号在 Python 中做了什么 [重复]

Question

我遇到了以下代码：

# O(n) space       
def rotate(self, nums, k):
    deque = collections.deque(nums)
    k %= len(nums)
    for _ in xrange(k):
        deque.appendleft(deque.pop())
    nums[:] = list(deque) # <- Code in question

nums[:] = 做了哪些 nums = 不做的事情？就此而言，nums[:] 做了哪些 nums 不做的事情？

Answer 1

nums = foo 重新绑定名称 nums 以引用 foo 所引用的同一对象。

nums[:] = foo 对nums 引用的对象调用切片分配，从而使原始对象的内容成为foo 内容的副本。

试试这个：

>>> a = [1,2]
>>> b = [3,4,5]
>>> c = a
>>> c = b
>>> print(a)
[1, 2]
>>> c = a
>>> c[:] = b
>>> print(a)
[3, 4, 5]

Answer 2

此语法是切片赋值。 [:] 的一部分表示整个列表。 nums[:] = 和 nums = 之间的区别在于后者不会替换原始列表中的元素。当有两个对列表的引用时，这是可以观察到的

>>> original = [1, 2, 3]
>>> other = original
>>> original[:] = [0, 0] # changes the contents of the list that both
                         # original and other refer to 
>>> other # see below, now you can see the change through other
[0, 0]

要查看差异，只需从上面的作业中删除 [:]。

>>> original = [1, 2, 3]
>>> other = original
>>> original = [0, 0] # original now refers to a different list than other
>>> other # other remains the same
[1, 2, 3]

---

从字面上看问题的标题，如果 list 是变量名而不是内置变量名，它将用省略号替换序列的长度

>>> list = [1,2,3,4]
>>> list[:] = [...]
>>> list
[Ellipsis]