Why does foo.append(bar) affect all elements in a

Question

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• List of lists changes reflected across sublists unexpectedly 12 answers

I create a list of lists and want to append items to the individual lists, but when I try to append to one of the lists (a[0].append(2)), the item gets added to all lists.

a = []
b = [1]

a.append(b)
a.append(b)

a[0].append(2)
a[1].append(3)
print(a)

Gives: [[1, 2, 3], [1, 2, 3]]

Whereas I would expect: [[1, 2], [1, 3]]

Changing the way I construct the initial list of lists, making b a float instead of a list and putting the brackets inside .append(), gives me the desired output:

a = []
b = 1

a.append([b])
a.append([b])

a[0].append(2)
a[1].append(3)
print(a)

Gives: [[1, 2], [1, 3]]

But why? It is not intuitive that the result should be different. I know this has to do with there being multiple references to the same list, but I don't see where that is happening.

Answer 1

It is because the list contains references to objects. Your list doesn't contain [[1 2 3] [1 2 3]], it is [<reference to b> <reference to b>].

When you change the object (by appending something to b), you are changing the object itself, not the list that contains the object.

To get the effect you desire, your list a must contain copies of b rather than references to b. To copy a list you can use the range [:]. For example, :

>>> a=[]
>>> b=[1]
>>> a.append(b[:])
>>> a.append(b[:])
>>> a[0].append(2)
>>> a[1].append(3)
>>> print a
[[1, 2], [1, 3]]

Answer 2

The key is this part:

a.append(b)
a.append(b)

You are appending the same list twice, so both a[0] and a[1] are references to the same list.

In your second example, you are creating new lists each time you call append like a.append([b]), so they are separate objects that are initialized with the same float value.

Answer 3

In order to make a shallow copy of a list, the idiom is

a.append(b[:])

which when doubled will cause a to have two novel copies of the list b which will not give you the aliasing bug you report.