Creating a new dictionary variable in a class python

Question

If I create a class like below:

class Foo(object): 
     foo = {'useful_info': 'useful_info'}  
     def __init__(self, foo, bar): 
         self.foo['foo'] = foo 
         self.bar = bar

and I create an instances of it and print it's attributes as:

Foo1 = Foo('foo1', 'bar1')
print(Foo1.foo)
print(Foo1.bar)

I get the output:

Foo1 = Foo('foo1', 'bar1')
print(Foo1.foo)
print(Foo1.bar)

However, now if I create a new instance named Foo2 and print the attributes of both Foo1 and Foo2 as:

Foo2 = Foo('foo2', 'bar2')

print("Foo2.foo: ", Foo2.foo) 
print("Foo2.bar: ", Foo2.bar) 
print("Foo1.foo: ", Foo1.foo) 
print("Foo1.bar: ", Foo1.bar)

I get the output:

Foo2.foo:  {'foo': 'foo2'}
Foo2.bar:  bar2
Foo1.foo:  {'foo': 'foo2'}
Foo1.bar:  bar1

The string attributes bar have been set as I expect, but the dictionary foo has been updated for both instances of Foo.

I can get round this by declaring the dict foo in the init magic method as:

class Foo(object): 
     def __init__(self, foo, bar): 
         self.foo = {'useful_info': 'useful_info'} 
         self.foo['foo'] = foo 
         self.bar = bar

but now the dictionary foo containing 'useful_info' isn't accessible until the init function is called i.e. if I use the line: Foo.foo I get an AttributeError.

Is there any way to create a new dictionary containing some 'useful_infomation' so that the useful info is accessible before the class is initialised and every new instance of a class has a different instance of this dictionary.

Cory Kramer · Accepted Answer · 2020-02-25 13:52:21Z

1

You can make a copy.deepcopy of the dictionary within __init__ so each instance has a deep copy of the dictionary, instead of sharing the same foo dictionary object.

import copy

class Foo(object): 
    foo = {'useful_info': 'useful_info'}  
    def __init__(self, bar): 
        self.foo = copy.deepcopy(Foo.foo)
        self.bar = bar

edited Feb 25, 2020 at 13:52

answered Feb 25, 2020 at 13:42

Cory Kramer

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3 Comments

Matt Ellis Over a year ago

Thanks! Though I think the self.foo['foo']... line should be replaced with self.foo = copy.deepcopy(self.foo)

AniMir Over a year ago

But, why do they share the same foo object ?

Matt Ellis Over a year ago

The original question had the argument 'foo' being set as a value under the key 'foo' of the dict foo. I understand this probably wasn't the best naming system! To answer this while keeping the functionality of the original question the dict foo would have to be re-created in the init and in that new dict the key foo should be set to the argument foo. Though I agree this pretty much answers my question.

juanpa.arrivillaga · Accepted Answer · 2020-02-25 13:44:57Z

1

Just combine both:

class Foo:
    foo = {'useful_info': 'useful_info'}
    def __init__(self, foo, bar):
        self.foo = {'useful_info': 'useful_info'}
        self.foo['foo'] = foo
        self.bar = bar

answered Feb 25, 2020 at 13:44

juanpa.arrivillaga

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2 Comments

AniMir Over a year ago

Wouldn't foo['useful_info'] be overwritten by the call of __init()__ ? If it was used and modified before that call, wouldn't it be an issue ?

juanpa.arrivillaga Over a year ago

@AniMir the class variable would not be modified, because an instance variable with the same name is created the line above it... self.foo = {'useful_info': 'useful_info'}

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