list.insert() in 实际上在 python 中做了什么?

2022-01-09 00:00:00 python python-3.x list insert

问题描述

我有这样的代码:

squares = []
for value in range(1, 5):
    squares.insert(value+1,value**2)

print(squares)
print(squares[0])
print(len(squares))

输出是:

[1, 4, 9, 16]

1

4

因此，即使我要求 python 在索引2"处插入1"，它也会在第一个可用索引处插入.那么插入"是如何做出决定的呢?

So even if I ask python to insert '1' at index '2', it inserts at the first available index. So how does 'insert' makes the decision?

解决方案

来自 Python3文档:

list.insert(i, x)

在给定位置插入一个项目.首先参数是要插入的元素的索引，所以a.insert(0, x) 在列表的前面插入，而 a.insert(len(a),x) 等价于 a.append(x).

Insert an item at a given position. The first argument is the index of the element before which to insert, so a.insert(0, x) inserts at the front of the list, and a.insert(len(a), x) is equivalent to a.append(x).

没有提到的是你可以给出一个超出范围的索引，然后 Python 会追加到列表中.

What is not mentionned is that you can give an index that is out of range and Python will then append to the list.

如果您深入研究 Python 实现，您会发现执行插入的 ins1 函数中的以下内容:

If you dig into the Python implementation you find the following in the ins1 function that does the insertion:

if (where > n)
    where = n;

所以基本上 Python 会将您的索引最大化到列表的长度.

So basically Python will max out your index to the length of the list.

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