DEV Community

Kelvin Wangonya
Kelvin Wangonya

Posted on

Shallow & deep copying in python

What happens when a variable is assigned to another variable in python? For example:

>>> x = 5
>>> y = x
Enter fullscreen mode Exit fullscreen mode

Both x and y will have the value 5. But, when x was assigned to y, y was not created as a completely new/separate object. Instead, an alias for x was created. That is, y points to the memory location of x. It does not have it's own memory location - yet.

>>> id(x)
140428600776960
>>> id(y)
140428600776960
>>> x is y
True
Enter fullscreen mode Exit fullscreen mode

You may never have any problems with this when working with immutable types because the alias is broken as soon as either of the two variables change.

>>> x += 2
>>> x
7
>>> y
5
>>> id(x)
140539682924864
>>> id(y)
140428600776960
>>> x is y
False
Enter fullscreen mode Exit fullscreen mode

But when working with mutable types, the alias is not broken when the original is updated. This means changes in x would reflect in y.

>>> x = [1,2,3]
>>> y = x
>>> x
[1, 2, 3]
>>> y
[1, 2, 3]
>>> x is y
True
>>> 
>>> x.append(4)
>>> x.append(5)
>>> x
[1, 2, 3, 4, 5]
>>> y
[1, 2, 3, 4, 5]
>>> x is y
True
Enter fullscreen mode Exit fullscreen mode

y was updated externally. This might be a cause of bugs if for example a value is updated by an external library and other variables are affected.

This can be prevented by creating shallow/deep copies of objects instead of using assignment.

Shallow copy

A shallow copy creates a new object, then populates it with references of the objects in the original object. Continuing with the previous example, a shallow copy can be created using either the list() or copy() command.

>>> z = list(x)
>>> z
[1, 2, 3, 4, 5]
Enter fullscreen mode Exit fullscreen mode

Now if some more values are appended to x, y will still be affected by z will not.

>>> x.append(6)
>>> x.append(7)
>>> x
[1, 2, 3, 4, 5, 6, 7]
>>> y
[1, 2, 3, 4, 5, 6, 7]
>>> z
[1, 2, 3, 4, 5]
Enter fullscreen mode Exit fullscreen mode

However, a shallow copy doesn't fully solve the problem because even though a new list was created, the objects in the list are still references to the objects in x.

As it is currently, updating x[0] would not affect z[0] because - immutable objects - the alias would be broken. But, if we were dealing with a list of lists, an update in x[0] would affect z[0].

>>> x = [[1,2], [3,4]]
>>> y = x
>>> z = list(x)
>>> x
[[1, 2], [3, 4]]
>>> y
[[1, 2], [3, 4]]
>>> z
[[1, 2], [3, 4]]
>>> x.append([5,6])
>>> x
[[1, 2], [3, 4], [5, 6]]
>>> y
[[1, 2], [3, 4], [5, 6]]
>>> z
[[1, 2], [3, 4]]
>>>
>>> x[0][1] = 'edited'
>>> x
[[1, 'edited'], [3, 4], [5, 6]]
>>> y
[[1, 'edited'], [3, 4], [5, 6]]
>>> z
[[1, 'edited'], [3, 4]]
Enter fullscreen mode Exit fullscreen mode

Deep copy

A deep copy creates a new object, and completely new instances of the objects in it. That is, a deep copied object is completely independent of the original. Updating objects in the original would not affect the deep copied object since there's no longer any connection.

>>> import copy
>>> x = [[1,2], [3,4]]
>>> z = copy.deepcopy(x)
>>> x
[[1, 2], [3, 4]]
>>> z
[[1, 2], [3, 4]]
>>> x is z
False
>>> x[0][1] = 'edited'
>>> x
[[1, 'edited'], [3, 4]]
>>> z
[[1, 2], [3, 4]]
Enter fullscreen mode Exit fullscreen mode

Latest comments (2)

Collapse
 
victorzakharenko profile image
VictorZakharenko

very nice! thx

Collapse
 
erikwhiting88 profile image
Erik

excellent write up, thanks for this article