Lists of lists: why does changing value at a[1][0] also change a[0][0]?

Question

Can anyone please explain what is happening here? I understand python ints are assigned by value (and not by reference).

>>> alist = [1,2,3]
>>> a = [[0]*3]*4
>>> a[0][0] = len(alist)
>>> alist.append(1)
>>> a[1][0] = len(alist)
>>> a[0][0]a
4
>>>

Off topic: Naming your variables as reserved words (*list*) is a recipe for disaster **:-)** — Savir, Mar 25 '13 at 22:31
There are less *different* list objects than are expected: `[] * x` *does not* preform a "deep copy". — , Mar 25 '13 at 22:32
The problem is that if I edit a[1][0] I don't expect a[0][0] to take the same value. — Jules Testard, Mar 25 '13 at 22:35

score 4 · Accepted Answer · answered Mar 25 '13 at 22:36

When you create a list using the * notation, you're repeating the same element. Not copies of the element, but the same exact object.

When the object you're repeating is immutable, such as an integer, you won't notice.

When the object is mutable, such as a list, you get the same list repeated. If you make a change to one list you're making it to all of them.

In your case you made a change to element [1] and it was reflected in element [0].

score 3 · Answer 2 · answered Mar 25 '13 at 22:35

3

The problem is that your outer list contains multiple copies of the same inner list value. Here's how you can tell:

a = [[0]*3]*4
print(a[0] is a[1]) # prints True

If you want a two dimensional list of zeros, I suggest using a list comprehension for the outer part so that you get separate inner lists instances:

a = [[0]*3 for _ in range(4)]
print(a[0] is a[1]) # prints False

a[0][0]=1
print(a[1][0]) # prints 0

answered Mar 25 '13 at 22:35

Blckknght

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Ok that explains everything! I must say this syntax is rather confusing. – Jules Testard Mar 25 '13 at 22:39

Jason Sperske · Answer 3 · 2013-03-25T22:41:14.877

2

This might help (just dumping the values of list and a at each line):

#With some commentary
>>> alist = [1,2,3] #Created a list with 3 elements
>>> alist
[1, 2, 3]
>>> a = [[0]*3]*4 #Created a list with 3 copies of one element and then copied it 4 times
>>> a
[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> a[0][0] = len(alist) #Set the 0th element of the 0th copy to 3
>>> a #Now you can see where that value was copied
[[3, 0, 0], [3, 0, 0], [3, 0, 0], [3, 0, 0]]
#The rest is just a permutation of that first 'bug'
>>> alist.append(1)
>>> alist
[1, 2, 3, 1]
>>> a[1][0] = len(alist)
>>> a
[[4, 0, 0], [4, 0, 0], [4, 0, 0], [4, 0, 0]]
>>> a[0][0]
4

An unintuitive thing about this code might be that [0]*3 seems do do what you expected while [[0]*3]*4 did not. For this I can only guess (not a Python guru yet) that a single length list of ints was handled in a way that seemed intuitive, while a list of lists fell back to a copy approach.

edited Mar 25 '13 at 22:41

answered Mar 25 '13 at 22:29

Jason Sperske

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Don't use names of built-ins for your variables, i.e. `list`. – Mark Ransom Mar 25 '13 at 22:31
+1 @MarkRansom. that too! – Jason Sperske Mar 25 '13 at 22:32
1

Mark, I think we'll all agree with you. But that comment is clearly better given to the OP. This poster is just trying to help the OP understand using their own terms. – john_science Mar 25 '13 at 22:32
I realize it was that way in the original question, but you should fix it in your answer. – Mark Ransom Mar 25 '13 at 22:32
well i do want to only access one element at a time. How should I do then? – Jules Testard Mar 25 '13 at 22:33
@JasonSperske This is the *start* of a good answer - however, spell out *what* is occurring instead of just providing the output that takes "some interpretation" to understand. – Mar 25 '13 at 22:35
Well it follows my thought process. I thought that a[x][y] was the regular way to access/write data in a 2D list in Python. I did not imagine this would be the behavior – Jules Testard Mar 25 '13 at 22:37
@JulesTestard, your approach to accessing is correct, it was your initialization of `a` that gave the odd behavior. – Jason Sperske Mar 25 '13 at 22:42

score 0 · Answer 4 · answered Mar 25 '13 at 22:35

No problem. [[0]*3]*4 created a list with 4 copies of a list of 3 elements. a[0], a[1], a[2] and a[3] all point to the same list.

>>> a=[[0]*3]*4
>>> a
[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> a[0][0]='hello'
>>> a
[['hello', 0, 0], ['hello', 0, 0], ['hello', 0, 0], ['hello', 0, 0]]
>>> for mylist in a:
    ...     print id(a)
    ...
    42701168
    42701168
    42701168
    42701168
>>>

Lists of lists: why does changing value at a[1][0] also change a[0][0]?

4 Answers4