Python lists confusion

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“Least Astonishment” in Python: The Mutable Default Argument
Python list confusion

I am a bit puzzled with lists of lists in Python. I have these two snippets:

li1 = [['a'], ['a'], ['a']]
print li1
for i in range(0, len(li1)):
    li1[i] += [i]
print li1

li2 = [['a']] * 3
print li2
for i in range(0, len(li2)):
    li2[i] += [i]
print li2

After creation li1 and li2 are the same, but when I add elements they behave differently:

[['a'], ['a'], ['a']]
[['a', 0], ['a', 1], ['a', 2]]
[['a'], ['a'], ['a']]
[['a', 0, 1, 2], ['a', 0, 1, 2], ['a', 0, 1, 2]]

Could some one please explain where the trick is?

I tried to look for similar questions but could actually formulate the problem, so asked. — Sergei G, Aug 24 '12 at 11:29
And you didn't see the question that had practically the *exact same title* as the one you're asking? — Tim Pietzcker, Aug 24 '12 at 11:31

Simeon Visser · Accepted Answer · 2012-08-24T14:28:45.340

In li2 = [['a']] * 3 you create one list with three list elements but these lists are actually the same object. That means: when you modify li2[0], you also modify li2[1] and li2[2].

The following line actually creates a list with three different list objects inside it:

li1 = [['a'], ['a'], ['a']]

In this case, when you modify li1[0] you only modify that list. The other lists are unaffected. This explains why you are getting different lists in li1 and li2.

score 1 · Answer 2 · answered Aug 24 '12 at 11:28

check the value of id() for each element,which clearly suggest the reason of their behavior.

>>> li1 = [['a'], ['a'], ['a']]
>>> for x in li1:                  #different id()
    id(x)


145497484
145514156
145511500

Same values of id():

>>> li1=['a']*3
>>> for x in li1:
    print id(x)


3078093024
3078093024
3078093024

Python lists confusion

2 Answers2