Python – How to avoid refer multiple names to one dict

Question

I am struggling with python dictionary. I have two dics named defaultSettings and personalSettings. I want to make function sets personalSettings‘s values to defaultSettings‘s values. But I don’t want to all changes be applied to defaulSettings.

I tried this code:

defaultSettings = {'s1': 1, 's2':2, 's3':3}

personalSettings = {'s1': 3, 's2':2, 's3':1}

print(defaultSettings)
print(personalSettings)
print('-*-*-*-*-*-*-*-*-*-*-*-*')

personalSettings = defaultSettings

print(defaultSettings)
print(personalSettings)
print('-*-*-*-*-*-*-*-*-*-*-*-*')

personalSettings['s1'] = 5

print(defaultSettings)
print(personalSettings)

And my output is:

{'s1': 1, 's2': 2, 's3': 3}
{'s1': 3, 's2': 2, 's3': 1}
-*-*-*-*-*-*-*-*-*-*-*-*
{'s1': 1, 's2': 2, 's3': 3}
{'s1': 1, 's2': 2, 's3': 3}
-*-*-*-*-*-*-*-*-*-*-*-*
{'s1': 5, 's2': 2, 's3': 3}
{'s1': 5, 's2': 2, 's3': 3}
-*-*-*-*-*-*-*-*-*-*-*-*

If I change a value of personalSettings after personalSettings = defaultSettings, defaultSettings‘s value changes too. I know this;

But I don’t know how avoid this or other way.

Asked By: Seminet

||

Source

Answer 1

Consider

>>> defaults = {'foo': 'bar', 'bee': 'blah'}
>>> actual = dict(defaults)
>>> actual['foo'] = 'BAZ'
>>> defaults
{'foo': 'bar', 'bee': 'blah'}
>>> actual
{'foo': 'BAZ', 'bee': 'blah'}

The trick is actual = dict(defaults). This way actual can be changed without affecting defaults.

If your dicts have sub-objects, consider deepcopy or json.dumps/loads to make a deep copy:

>>> myDict = {'s1':1,  's2':{'b1':2, 'b2':3}}
>>>
>>> from copy import deepcopy
>>>
>>> myDict2 = deepcopy(myDict)
>>>
>>> myDict2['s2']['b1'] = 99
>>>
>>> myDict
{'s1': 1, 's2': {'b1': 2, 'b2': 3}}
>>> myDict2
{'s1': 1, 's2': {'b1': 99, 'b2': 3}}

Answered By: gog

Answer 2

I would use defaultSettings.copy() or dict(defaultSettings) to create a copy of the dict

Answered By: David Tejuosho

Python – How to avoid refer multiple names to one dict

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