正如上面其他人所述，列表(当相当大时)是通过分配固定数量的空间来实现的，如果该空间应该被填满，则分配更大数量的空间并复制元素。

To understand why the method is O(1) amortized, without loss of generality, assume we have inserted a = 2^n elements, and we now have to double our table to 2^(n+1) size. That means we're currently doing 2^(n+1) operations. Last copy, we did 2^n operations. Before that we did 2^(n-1)... all the way down to 8,4,2,1. Now, if we add these up, we get 1 + 2 + 4 + 8 + ... + 2^(n+1) = 2^(n+2) - 1 < 4*2^n = O(2^n) = O(a) total insertions (i.e. O(1) amortized time). Also, it should be noted that if the table allows deletions the table shrinking has to be done at a different factor (e.g 3x)

根据文档，

Python的列表实际上是变长数组，而不是lisp风格的链表。

这取决于实现，但IIRC:

CPython使用指针数组 Jython使用数组列表 IronPython显然也使用数组。您可以浏览源代码来找到答案。

因此它们都有O(1)个随机访问。

正如上面其他人所述，列表(当相当大时)是通过分配固定数量的空间来实现的，如果该空间应该被填满，则分配更大数量的空间并复制元素。

To understand why the method is O(1) amortized, without loss of generality, assume we have inserted a = 2^n elements, and we now have to double our table to 2^(n+1) size. That means we're currently doing 2^(n+1) operations. Last copy, we did 2^n operations. Before that we did 2^(n-1)... all the way down to 8,4,2,1. Now, if we add these up, we get 1 + 2 + 4 + 8 + ... + 2^(n+1) = 2^(n+2) - 1 < 4*2^n = O(2^n) = O(a) total insertions (i.e. O(1) amortized time). Also, it should be noted that if the table allows deletions the table shrinking has to be done at a different factor (e.g 3x)

在CPython中，列表是指针的数组。Python的其他实现可以选择以不同的方式存储它们。

在CPython中，list是作为动态数组实现的，因此当我们追加时，不仅添加了一个宏，而且分配了更多的空间，这样每次都不应该添加新的空间。