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pythonarraysinitialization
Tanky Woo
Tanky Woo

Reputation: 5106

The initialization of 2-dimensional array in python

I asked a question about initializing a 2 dimensional array yesterday, this is the link: How to implement this C++ source in python?

There is a problem in the answer, a friend mentioned a way:

G = [[0]*11]*11

But in this way, when I change the G[0][0] to 2, all the G[i][0](0<=i<11) will all change to 2, but I don't know why?

Supplement:

This is what I thought: The 0 or other number is immutable, so we change one of them, the others will not be changed. But the list [0, 0 ,0 ,.....] is mutable, so when we [0, 0, ...] * 11, all the [0, 0, ...] list will be the same, as is function is True. am I right?

Upvotes: 1

Views: 570

Answers (2)

Mark Ransom
Mark Ransom

Reputation: 308111

The *11 notation makes 11 references to the same object. If the object is immutable you don't notice, because any attempt to change it changes the reference to a different object. When the object is mutable you can modify it, like assigning to a member of a list; since all the references are to the same object, all of them get modified at the same time.

Mutable/immutable might seem to change things, but it doesn't - Python is being consistent in both cases. Consider this example:

G[0] = [3]*11

You'll see that G[1] has not changed.

Upvotes: 2

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 798526

Because you have 11 references to the same list.

G = [[0] * 11 for x in range(11)]

Upvotes: 4

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