Python初始化列表列表 [英] Python initializing a list of lists

问题描述

可能重复:
Python列表附加行为

Possible Duplicate:
Python list append behavior

我打算初始化一个长度为n的列表列表.

I intend to initialize a list of list with length of n.

x = [[]] * n

但是，这以某种方式将列表链接在一起.

However, this somehow links the lists together.

>>> x = [[]] * 3
>>> x[1].append(0)
>>> x
[[0], [0], [0]]

我希望有类似的东西:

[[], [0], []]

有什么想法吗?

推荐答案

问题是它们在内存中的列表完全相同.使用[x]*n语法时，得到的是n许多x对象的列表，但是它们都是对同一对象的引用.它们不是不同的实例，而只是对同一实例的n引用.

The problem is that they're all the same exact list in memory. When you use the [x]*n syntax, what you get is a list of n many x objects, but they're all references to the same object. They're not distinct instances, rather, just n references to the same instance.

要列出3个不同的列表，请执行以下操作:

To make a list of 3 different lists, do this:

x = [[] for i in range(3)]

这为您提供了[]的3个单独实例，这就是您想要的

This gives you 3 separate instances of [], which is what you want

[[]]*n类似于

l = []
x = []
for i in range(n):
    x.append(l)

[[] for i in range(3)]类似于:

x = []
for i in range(n):
    x.append([])   # appending a new list!

---

In [20]: x = [[]] * 4

In [21]: [id(i) for i in x]
Out[21]: [164363948, 164363948, 164363948, 164363948] # same id()'s for each list,i.e same object


In [22]: x=[[] for i in range(4)]

In [23]: [id(i) for i in x]
Out[23]: [164382060, 164364140, 164363628, 164381292] #different id(), i.e unique objects this time

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