[FIXED] Why does foo.append(bar) affect all elements in a list of lists?

Issue

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.

Solution

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]]

Answered By - Bryan Oakley

Answer Checked By - Timothy Miller (PHPFixing Admin)