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。

Python中的列表类似于数组，可以存储多个值。列表是可变的，这意味着你可以改变它。你应该知道的更重要的事情是，当我们创建一个列表时，Python会自动为该列表变量创建一个reference_id。如果你通过分配其他变量来改变它，主列表也会改变。让我们举个例子:

list_one = [1,2,3,4]

my_list = list_one

#my_list: [1,2,3,4]

my_list.append("new")

#my_list: [1,2,3,4,'new']
#list_one: [1,2,3,4,'new']

我们添加了my_list，但是主列表已经改变了。那意味着列表没有赋值为一个复制列表赋值为它的引用。

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。

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。

我推荐Laurent Luce的文章“Python列表实现”。对我来说真的很有用，因为作者解释了如何在CPython中实现列表，并为此使用了出色的图表。

List object C structure A list object in CPython is represented by the following C structure. ob_item is a list of pointers to the list elements. allocated is the number of slots allocated in memory. typedef struct { PyObject_VAR_HEAD PyObject **ob_item; Py_ssize_t allocated; } PyListObject; It is important to notice the difference between allocated slots and the size of the list. The size of a list is the same as len(l). The number of allocated slots is what has been allocated in memory. Often, you will see that allocated can be greater than size. This is to avoid needing calling realloc each time a new elements is appended to the list.

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附加 我们在列表中添加一个整数:l.p append(1)。会发生什么呢? 我们继续添加一个元素:l.p append(2)。当n+1 = 2时调用List_resize，但因为分配的大小是4，所以不需要分配更多的内存。当我们再加两个整数l.append(3)， l.append(4)时也会发生同样的事情。下面的图表显示了我们目前所拥有的。

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插入 让我们在位置1插入一个新的整数(5):l.insert(1,5)，看看内部发生了什么。

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Pop When you pop the last element: l.pop(), listpop() is called. list_resize is called inside listpop() and if the new size is less than half of the allocated size then the list is shrunk. You can observe that slot 4 still points to the integer but the important thing is the size of the list which is now 4. Let’s pop one more element. In list_resize(), size – 1 = 4 – 1 = 3 is less than half of the allocated slots so the list is shrunk to 6 slots and the new size of the list is now 3. You can observe that slot 3 and 4 still point to some integers but the important thing is the size of the list which is now 3.

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删除 Python列表对象有一个删除特定元素的方法:l.remove(5)。

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)