python - Understanding dict.copy() - shallow or deep?

474
votes

While reading up the documentation for dict.copy(), it says that it makes a shallow copy of the dictionary. Same goes for the book I am following (Beazley's Python Reference), which says:

The m.copy() method makes a shallow copy of the items contained in a mapping object and places them in a new mapping object.

Consider this:

>>> original = dict(a=1, b=2)
>>> new = original.copy()
>>> new.update({'c': 3})
>>> original
{'a': 1, 'b': 2}
>>> new
{'a': 1, 'c': 3, 'b': 2}

So I assumed this would update the value of original (and add 'c': 3) also since I was doing a shallow copy. Like if you do it for a list:

>>> original = [1, 2, 3]
>>> new = original
>>> new.append(4)
>>> new, original
([1, 2, 3, 4], [1, 2, 3, 4])

This works as expected.

Since both are shallow copies, why is that the dict.copy() doesn't work as I expect it to? Or my understanding of shallow vs deep copying is flawed?

7
pythondictionarycopy
Quaint that they don't explain "shallow". Insider knowledge, wink. Just the dict and keys are a copy while nested dicts inside that first level are references, cannot be deleted in a loop for example. Thus Python's dict.copy() in that case is neither useful nor intuitive. Thanks for your question. – gseattle

7 Answers

1103
votes

By "shallow copying" it means the content of the dictionary is not copied by value, but just creating a new reference.

>>> a = {1: [1,2,3]}
>>> b = a.copy()
>>> a, b
({1: [1, 2, 3]}, {1: [1, 2, 3]})
>>> a[1].append(4)
>>> a, b
({1: [1, 2, 3, 4]}, {1: [1, 2, 3, 4]})

In contrast, a deep copy will copy all contents by value.

>>> import copy
>>> c = copy.deepcopy(a)
>>> a, c
({1: [1, 2, 3, 4]}, {1: [1, 2, 3, 4]})
>>> a[1].append(5)
>>> a, c
({1: [1, 2, 3, 4, 5]}, {1: [1, 2, 3, 4]})

So:

  1. b = a: Reference assignment, Make a and b points to the same object.

    Illustration of 'a = b': 'a' and 'b' both point to '{1: L}', 'L' points to '[1, 2, 3]'.

  2. b = a.copy(): Shallow copying, a and b will become two isolated objects, but their contents still share the same reference

    Illustration of 'b = a.copy()': 'a' points to '{1: L}', 'b' points to '{1: M}', 'L' and 'M' both point to '[1, 2, 3]'.

  3. b = copy.deepcopy(a): Deep copying, a and b's structure and content become completely isolated.

    Illustration of 'b = copy.deepcopy(a)': 'a' points to '{1: L}', 'L' points to '[1, 2, 3]'; 'b' points to '{1: M}', 'M' points to a different instance of '[1, 2, 3]'.

39
votes

It's not a matter of deep copy or shallow copy, none of what you're doing is deep copy.

Here:

>>> new = original

you're creating a new reference to the the list/dict referenced by original.

while here:

>>> new = original.copy()
>>> # or
>>> new = list(original) # dict(original)

you're creating a new list/dict which is filled with a copy of the references of objects contained in the original container.

38
votes

Take this example:

original = dict(a=1, b=2, c=dict(d=4, e=5))
new = original.copy()

Now let's change a value in the 'shallow' (first) level:

new['a'] = 10
# new = {'a': 10, 'b': 2, 'c': {'d': 4, 'e': 5}}
# original = {'a': 1, 'b': 2, 'c': {'d': 4, 'e': 5}}
# no change in original, since ['a'] is an immutable integer

Now let's change a value one level deeper:

new['c']['d'] = 40
# new = {'a': 10, 'b': 2, 'c': {'d': 40, 'e': 5}}
# original = {'a': 1, 'b': 2, 'c': {'d': 40, 'e': 5}}
# new['c'] points to the same original['d'] mutable dictionary, so it will be changed
8
votes

Adding to kennytm's answer. When you do a shallow copy parent.copy() a new dictionary is created with same keys,but the values are not copied they are referenced.If you add a new value to parent_copy it won't effect parent because parent_copy is a new dictionary not reference.

parent = {1: [1,2,3]}
parent_copy = parent.copy()
parent_reference = parent

print id(parent),id(parent_copy),id(parent_reference)
#140690938288400 140690938290536 140690938288400

print id(parent[1]),id(parent_copy[1]),id(parent_reference[1])
#140690938137128 140690938137128 140690938137128

parent_copy[1].append(4)
parent_copy[2] = ['new']

print parent, parent_copy, parent_reference
#{1: [1, 2, 3, 4]} {1: [1, 2, 3, 4], 2: ['new']} {1: [1, 2, 3, 4]}

The hash(id) value of parent[1], parent_copy[1] are identical which implies [1,2,3] of parent[1] and parent_copy[1] stored at id 140690938288400.

But hash of parent and parent_copy are different which implies They are different dictionaries and parent_copy is a new dictionary having values reference to values of parent

5
votes

"new" and "original" are different dicts, that's why you can update just one of them.. The items are shallow-copied, not the dict itself.

2
votes

Contents are shallow copied.

So if the original dict contains a list or another dictionary, modifying one them in the original or its shallow copy will modify them (the list or the dict) in the other.

1
votes

In your second part, you should use new = original.copy()

.copy and = are different things.