Python 初始化列表列表
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Python initializing a list of lists
提问by Amir
Possible Duplicate:
Python list append behavior
可能重复：
Python 列表追加行为
I intend to initialize a list of list with length of n.
我打算初始化一个长度为 n 的列表列表。
x = [[]] * n
However, this somehow links the lists together.
但是，这以某种方式将列表链接在一起。
>>> x = [[]] * 3
>>> x[1].append(0)
>>> x
[[0], [0], [0]]
I expect to have something like:
我希望有类似的东西：
[[], [0], []]
Any ideas?
有任何想法吗？
采纳答案by inspectorG4dget
The problem is that they're all the same exact list in memory. When you use the [x]*nsyntax, what you get is a list of nmany xobjects, but they're all references to the same object. They're not distinct instances, rather, just nreferences to the same instance.
问题是它们在内存中都是完全相同的列表。当您使用该[x]*n语法时，您得到的是n许多x对象的列表，但它们都是对同一对象的引用。它们不是不同的实例，而是n对同一个实例的引用。
To make a list of 3 different lists, do this:
要制作 3 个不同列表的列表，请执行以下操作：
x = [[] for i in range(3)]
This gives you 3 separate instances of [], which is what you want
这为您提供了 3 个单独的 实例[]，这正是您想要的
[[]]*nis similar to
[[]]*n类似于
l = []
x = []
for i in range(n):
x.append(l)
While [[] for i in range(3)]is similar to:
虽然[[] 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
