Python List Mutable

I am trying to use Python term to explain why the following happens, can somebody explain why tmp becomes to [[1,2,3]] not stay as [[1,2]]? arr = [] tmp = [1,2] arr.append(tmp) pri

Solution 1:

arr = [] is an empty list, and when you append tmp to it via:

tmp = [1, 2]
arr.append(tmp)

You are putting tmp in the arr list, thus giving you arr = [tmp] which can be expanded to arr = [[1,2]]. But the neat thing here is that you maintain a reference to to the list, [1,2] via the temp variable. Thus, when you append temp you are appending the same list that is in arr.

For a bit further clarification, just because you are appending tmp to arr doesn't mean that the resulting list [[1,2]] is all going to be one continuous block in memory. You are going to have the arr list and the first element of arr is going to be a pointer to the list tmp.

Solution 2:

All the comments are great ones.

arr.append(tmp)
print arr # [[1,2]]

As you can see, the result is NOT:

print arr # [1,2]

So, arr just holds the reference to tmp array. If my guess is write you are looking for:

 arr.extend(tmp)
 print arr # [1,2]

More on difference between append vs. extend list methods in python

Solution 3:

That's because of both tmp and arr[0] points to the same object. Just check it here, step by step:

http://www.pythontutor.com/visualize.html

First print statement Second print statement

You can manually check it by using id built-in

>>> arr = []
>>> tmp = [1,2]
>>> arr.append(tmp)
>>> id(tmp)
4404123192
>>> id(arr[0])
4404123192
>>> assert id(tmp) == id(arr[0])
>>> tmp.append(3) # allocate more memory (if needs) and add '3' to object (list) with id 4404123192
>>> id(tmp)
4404123192
>>> id(arr[0])
4404123192
>>> print arr
[[1, 2, 3]]