Python managing multiple dictionaries

Question

I am doing some very basic operations with Python (3.6, but also tested in Python 2.11 - behavior is the same) and dictionaries. Here is the code sample:

max_values = dict()
min_values = dict()

initial_values = {"stat": 2, "value": 5.5}

max_values["zzz"] = initial_values
min_values["zzz"] = initial_values

So far so good, when I print those dictionaries:

for key, value in max_values.items():
    print("[%s] Key [%s] -> Value [%s]" % ("max_values", key, value))

for key, value in min_values.items():
    print("[%s] Key [%s] -> Value [%s]" % ("min_values", key, value))

I get my expected results:

[max_values] Key [zzz] -> Value [{'stat': 2, 'value': 5.5}]
[min_values] Key [zzz] -> Value [{'stat': 2, 'value': 5.5}]

But if I attempt to modify with the values, for example:

min_values["zzz"]["value"] = 1.0

And do the same print again, I get:

[max_values] Key [zzz] -> Value [{'stat': 2, 'value': 1.0}]
[min_values] Key [zzz] -> Value [{'stat': 2, 'value': 1.0}]

Question is: Why was the max_values dictionary changed? I have only modified min_values, but this new key assignment for min_value dictionary also modified max_values. Am I doing something wrong?

Assingment **_does not copy_**. Both of the dictionaries key's have a reference to the same dictionary. You need to explicitly tell Python to make a copy of `initial_values` for both keys. — Christian Dean, Jul 17 '17 at 19:34
@Jean-FrançoisFabre Or `max_values["zzz"] = initial_values.copy()`. — Christian Dean, Jul 17 '17 at 19:36
Perhaps [this](https://stackoverflow.com/questions/19951816/python-changes-to-my-copy-variable-affect-the-original-variable) would work. It's not an exact duplicate, but it explains what assignment does in Python. — Christian Dean, Jul 17 '17 at 19:40
@ChristianDean you are correct, it's intended functionality in Python (and I assume in other languages as well). For anyone else willing to read a bit more about the difference between assignment and copying this link can help you: https://docs.python.org/3/library/copy.html — buuren, Jul 17 '17 at 21:39

score 1 · Answer 1 · answered Jul 17 '17 at 19:35

1

It's because the max_values and min_values are linked together. That is, they are both referencing initial_values. So if you change one you change the other. One solution would be to use deepcopy.

answered Jul 17 '17 at 19:35

Cary Shindell

1,336
8
25

Mr_U4913 · Answer 2 · 2017-07-17T20:00:14.403

0

As you can see from the graph, Max_value and Min_value reference to Initial_value , which means Max_value and Min_value link to the address where you create initial_value dictionary. Once you update the value inside of the initial_val, it will also show in both Max_value and Min_value.

To avoid this problem, you can use from copy import deepcopy.

my_dict = {'1': 1, '1': 2}
new_dict= deepcopy(my_dict)

In this way, it creates a new instance for your old dictionary.

edited Jul 17 '17 at 20:00

answered Jul 17 '17 at 19:54

Mr_U4913

1,294
8
12

That was the case! Thanks for the visual representation. – buuren Jul 17 '17 at 21:41

score 0 · Accepted Answer · answered Jul 17 '17 at 20:00

You have only one dict initial_values and both of your dict are refers to the same dict initial_values. You have to copy this dict to get another dict.

Try to:

max_values["zzz"] = initial_values
min_values["zzz"] = dict(initial_values)

or:

max_values["zzz"] = initial_values
min_values["zzz"] = initial_values.copy()

And now this constructions:

min_values["zzz"]["value"] = 1.0

for key, value in max_values.items():
    print("[%s] Key [%s] -> Value [%s]" % ("max_values", key, value))

for key, value in min_values.items():
    print("[%s] Key [%s] -> Value [%s]" % ("min_values", key, value))

Get expected result:

[max_values] Key [zzz] -> Value [{'stat': 2, 'value': 5.5}]
[min_values] Key [zzz] -> Value [{'stat': 2, 'value': 1.0}]

score -2 · Answer 4 · edited Jul 17 '17 at 19:37

-2

Because you gave both of the values same dictionary just look at the first two lines of your code.

edited Jul 17 '17 at 19:37

Christian Dean

22,138
7
54
87

answered Jul 17 '17 at 19:36

NapolyoN

337
3
12

Python managing multiple dictionaries

4 Answers4