why does the copy of the original Dictionary change when appending the original and deleting the original?

Question

so basically i know that shallow copy and deep copy varies. what i am trying to understand is that when i make a direct copy:

d1 = {"a": "1"}
d2 = {}
d2 = d1
d1["b"] = "2"
del d1

print(d2)

this returns the output as

{"a": "1", "b": "2"}

my question is why does this happen, and what can i do to not let this happen? i dont want d2 to change once i make the copy. Is there any way to do it without using copy.deepcopy()?

you never make a copy anywhere in the above code, neither shallow nor deep. — juanpa.arrivillaga, Jul 08 '20 at 01:07
You really should read the following: https://nedbatchelder.com/text/names.html — juanpa.arrivillaga, Jul 08 '20 at 01:10
The linked duplicate is about lists, but the same principles apply. `dict` objects have a `copy` method if you want to copy them, or you can just do `d2 = dict(d1)`. Note, `d2 = {}` before that is completely pointless. — juanpa.arrivillaga, Jul 08 '20 at 01:16

Addy · Answer 1 · 2020-07-08T01:23:02.290

1

No you cannot copy a object like that,When you are saying d2=d1 you are not making a copy of d1, it results in d2 being just another name for d1. If you don't want it to update "b":"2" use dict(d1)

d1={"a":1}
d2=dict(d1)
d1["b"]="2"
del d1
print(d2)

edited Jul 08 '20 at 01:23

answered Jul 08 '20 at 01:15

Addy

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score 0 · Answer 2 · answered Jul 08 '20 at 01:12

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Use .copy() to create a (shallow copy) of d1 like so:

d1 = {"a": "1"}
d2 = d1.copy()
d1["a"] = "2"
print(d2)

which outputs:

{'a': '1'}

The reason the original code did not work as you expected is that both d1 and d2 were pointing at the same object.

answered Jul 08 '20 at 01:12

C. Pappy

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score 0 · Answer 3 · answered Jul 08 '20 at 01:12

0

By assigning d2 = d1 they both point to the same variable in memory. You can modify this value with either name as they both point to the same thing.

This is not a copy you are only incrementing the reference count by one. By using del d1 you are decrementing the reference count by one but d2 still points to the same value in memory.

To create a copy where they are both separate entities then use d2 = d1.copy().

answered Jul 08 '20 at 01:12

Jab

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so if i delete both the references d1 and d2, what happens to that memory location? does it still store the same dictionary, but without any reference to any other object there? is that what we generally term as garbage values? – Metison Jul 10 '20 at 20:21
There is a garbage collector that runs in the background and clears up objects with no reference count. – Jab Jul 12 '20 at 21:19

why does the copy of the original Dictionary change when appending the original and deleting the original?

3 Answers3