Why updating “shallow” copy dictionary doesn’t update “original” dictionary? [duplicate]

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]'.

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