python operator "*" with List， any others like it?

Question

When using * to create list like:

>>> ll = [None] * 3
>>> ll
[None, None, None]
>>> ll[0] = 2
>>> ll
[2, None, None]

And then using two dimension,

>>> ll2 = 3 *[ 3 * [None]]
>>> ll2
[[None, None, None], [None, None, None], [None, None, None]]
>>> ll2[0][0] = 2
>>> ll2
[[2, None, None], [2, None, None], [2, None, None]]

So, create list by operator *, if the element is object like list. It doesn't create new object, just use the referenced object.

Are there any other operators like this in Python?

Edit:

Sorry, My representation is bad. What I want to know is, are there any other operators using like "*" will not create multiple reference.

To all the answerers - I think OP's asking about: *Are there any other operators like this in Python?* **Not:** *What's wrong with my code and how can I fix it?* — Remi Guan, Mar 21 '16 at 05:39
@Rogalski That problem is what I know. And What I ask is, are there any other operators using like "*" will not create multiple reference. — Yanhui Zhou, Mar 21 '16 at 07:40
[How to define two-dimensional array in python](http://stackoverflow.com/questions/6667201/how-to-define-two-dimensional-array-in-python), [How to initialize a two-dimensional array in Python?](http://stackoverflow.com/questions/2397141/how-to-initialize-a-two-dimensional-array-in-python) — GingerPlusPlus, Mar 21 '16 at 08:31

score 6 · Accepted Answer · answered Mar 21 '16 at 05:26

6

If you multiply the list of lists, it will create multiple reference of the inner lists.

>>> lst = 3 * [3 * [None]]
>>> lst[0] is lst[1]
True
>>> lst[0] is lst[2]
True

You need to create separated lists. For example, using list comprehension:

>>> lst = [3 * [None] for i in range(3)]
>>> lst[0][0] = 2
>>> lst
[[2, None, None], [None, None, None], [None, None, None]]

answered Mar 21 '16 at 05:26

falsetru

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I know that, what I want to know is that, are there any other operators using like "*" will not create multiple reference. – Yanhui Zhou Mar 21 '16 at 07:27
@YanhuiZhou, No operator that does what you want exists as far as I know. – falsetru Mar 21 '16 at 07:27
OK, thanks. When I found this one, I desire to know to is there any others. And I can not search anything about it. – Yanhui Zhou Mar 21 '16 at 07:31

score 2 · Answer 2 · answered Mar 21 '16 at 05:27

This has nothing to do with the * operator, but with the fact that you are using a list. [3 * [None]] creates a single list object, and you store a list of 3 references to this list in ll2. If you change one of the referenced lists, you change all the others, too. This behaviour will happen with all mutable objects

score 1 · Answer 3 · answered Mar 21 '16 at 05:25

This has nothing to do with the use of the * operator for multiplying a list. The assignment is by reference or in other words l is a mutable alias. To demonstrate the point, the following code will do the same thing.

l = [None, None, None]
l12 = [l, l, l]
print l12
>>> [[None, None, None], [None, None, None], [None, None, None]]
l[0]=2
print l12
>>> [[2, None, None], [2, None, None], [2, None, None]]

score 0 · Answer 4 · answered Mar 21 '16 at 07:35

0

list comprehension

>>> ll=[[None for r in range(3)] for k in range(3)]
>>> ll[0][0]=3
>>> ll
[[3, None, None], [None, None, None], [None, None, None]]

answered Mar 21 '16 at 07:35

python必须死

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python operator "*" with List， any others like it?

Edit:

4 Answers4