Copying python lists

Question

Just when I thought I had understood how Python lists work...

>>> a = [1,2,3]
>>> b = a[:]
>>> b
[1,2,3]
>>> b[1]=100
>>> b
[1,100,3]
>>> a
[1,2,3]

So far,so good. I am initializing b with the contents of a, so that b points to a different object. As a consequence, changes in b don't affect a.

Now take a look at this other example:

>>> a = [[1,2,3],[4,5,6],[7,8,9]]
>>> b = a[:][:]
>>> b
[[1,2,3],[4,5,6],[7,8,9]]
>>> b[1][1] = 100
>>> b
[[1,2,3],[4,100,6],[7,8,9]]
>>> a
[[1,2,3],[4,100,6],[7,8,9]]

Why has the change in b affected a this time? What is different from the previous example?

Answer 1

The slicing operation x[:] makes a shallow copy. That means, the outer list is different, but is contains the exact same elements. assume a = [[1]]:

b = a[:] # is the same as:
b = [x for x in a]

>>> a[0] is b[0]
True

The double slicing ([:][:]) is doing nothing more than that - again:

b = a[:][:] # is the same as:
b = [y for y in [x for x in a]]

>>> a[0] is b[0]
True

A shallow copy of a shallow copy is, well, a shallow copy.

So b is still a shallow copy of a - these lists are different objects, but they contain identical elements. Then you mutate the inner list in b, but it is the same list in a.

2-level shallow copy can be done with b=[x[:] for x in a]. It extends the same way for n-level copy.

By the way, the term "shallow copy" has the same meaning for any class or container.

If you want real deep copy, you should consider using deep copy.

Answer 2

The silcing [:] makes a shallow copy. but inner objects remain not copied so if they are mutable and You will modified them they will be modified ;) and in both list You will see this

use list comprehension

b = [i[:] for i in a]

or copy.deepcopy()

import copy
b = copy.deepcopy(a)