How to avoid having class data shared among instances?


What I want is this behavior:

class a:
    list = []

x = a()
y = a()


print(x.list) # prints [1, 3]
print(y.list) # prints [2, 4]

Of course, what really happens when I print is:

print(x.list) # prints [1, 2, 3, 4]
print(y.list) # prints [1, 2, 3, 4]

Clearly they are sharing the data in class a. How do I get separate instances to achieve the behavior I desire?

Asked By: 8steve8



You declared “list” as a “class level property” and not “instance level property”. In order to have properties scoped at the instance level, you need to initialize them through referencing with the “self” parameter in the __init__ method (or elsewhere depending on the situation).

You don’t strictly have to initialize the instance properties in the __init__ method but it makes for easier understanding.

Answered By: jldupont

You want this:

class a:
    def __init__(self):
        self.list = []

Declaring the variables inside the class declaration makes them “class” members and not instance members. Declaring them inside the __init__ method makes sure that a new instance of the members is created alongside every new instance of the object, which is the behavior you’re looking for.

Answered By: abyx

Yes you must declare in the “constructor” if you want that the list becomes an object property and not a class property.

Answered By: osanchezmon

The accepted answer works but a little more explanation does not hurt.

Class attributes do not become instance attributes when an instance is created. They become instance attributes when a value is assigned to them.

In the original code no value is assigned to list attribute after instantiation; so it remains a class attribute. Defining list inside __init__ works because __init__ is called after instantiation. Alternatively, this code would also produce the desired output:

>>> class a:
    list = []

>>> y = a()
>>> x = a()
>>> x.list = []
>>> y.list = []
>>> x.list.append(1)
>>> y.list.append(2)
>>> x.list.append(3)
>>> y.list.append(4)
>>> print(x.list)
[1, 3]
>>> print(y.list)
[2, 4]

However, the confusing scenario in the question will never happen to immutable objects such as numbers and strings, because their value cannot be changed without assignment. For example a code similar to the original with string attribute type works without any problem:

>>> class a:
    string = ''

>>> x = a()
>>> y = a()
>>> x.string += 'x'
>>> y.string += 'y'
>>> x.string
>>> y.string

So to summarize: class attributes become instance attributes if and only if a value is assigned to them after instantiation, being in the __init__ method or not. This is a good thing because this way you can have static attributes if you never assign a value to an attribute after instantiation.

Answered By: jurgenreza

So nearly every response here seems to miss a particular point. Class variables never become instance variables as demonstrated by the code below. By utilizing a metaclass to intercept variable assignment at the class level, we can see that when a.myattr is reassigned, the field assignment magic method on the class is not called. This is because the assignment creates a new instance variable. This behavior has absolutely nothing to do with the class variable as demonstrated by the second class which has no class variables and yet still allows field assignment.

class mymeta(type):
    def __init__(cls, name, bases, d):

    def __setattr__(cls, attr, value):
        print("setting " + attr)
        super(mymeta, cls).__setattr__(attr, value)

class myclass(object):
    __metaclass__ = mymeta
    myattr = []

a = myclass()
a.myattr = []           #NOTHING IS PRINTED
myclass.myattr = [5]    #change is printed here
b = myclass()
print(b.myattr)         #pass through lookup on the base class

class expando(object):

a = expando()
a.random = 5            #no class variable required
print(a.random)         #but it still works

IN SHORT Class variables have NOTHING to do with instance variables.

More clearly They just happen to be in the scope for lookups on instances. Class variables are in fact instance variables on the class object itself. You can also have metaclass variables if you want as well because metaclasses themselves are objects too. Everything is an object whether it is used to create other objects or not, so do not get bound up in the semantics of other languages usage of the word class. In python, a class is really just an object that is used to determine how to create other objects and what their behaviors will be. Metaclasses are classes that create classes, just to further illustrate this point.

Answered By: TheQabalist

Although the accepted anwer is spot on, I would like to add a bit description.

Let’s do a small exercise

first of all define a class as follows:

class A:
    temp = 'Skyharbor'

    def __init__(self, x):
        self.x = x

    def change(self, y):
        self.temp = y

So what do we have here?

  • We have a very simple class which has an attribute temp which is a string
  • An __init__ method which sets self.x
  • A change method sets self.temp

Pretty straight forward so far yeah? Now let’s start playing around with this class. Let’s initialize this class first:

a = A('Tesseract')

Now do the following:

>>> print(a.temp)
>>> print(A.temp)

Well, a.temp worked as expected but how the hell did A.temp work? Well it worked because temp is a class attribute. Everything in python is an object. Here A is also an object of class type. Thus the attribute temp is an attribute held by the A class and if you change the value of temp through A (and not through an instance of a), the changed value is going to be reflected in all the instance of A class.
Let’s go ahead and do that:

>>> A.temp = 'Monuments'
>>> print(A.temp)
>>> print(a.temp)

Interesting isn’t it? And note that id(a.temp) and id(A.temp) are still the same.

Any Python object is automatically given a __dict__ attribute, which contains its list of attributes. Let’s investigate what this dictionary contains for our example objects:

>>> print(A.__dict__)
    'change': <function change at 0x7f5e26fee6e0>,
    '__module__': '__main__',
    '__init__': <function __init__ at 0x7f5e26fee668>,
    'temp': 'Monuments',
    '__doc__': None
>>> print(a.__dict__)
{x: 'Tesseract'}

Note that temp attribute is listed among A class’s attributes while x is listed for the instance.

So how come that we get a defined value of a.temp if it is not even listed for the instance a. Well that’s the magic of __getattribute__() method. In Python the dotted syntax automatically invokes this method so when we write a.temp, Python executes a.__getattribute__('temp'). That method performs the attribute lookup action, i.e. finds the value of the attribute by looking in different places.

The standard implementation of __getattribute__() searches first the internal dictionary (dict) of an object, then the type of the object itself. In this case a.__getattribute__('temp') executes first a.__dict__['temp'] and then a.__class__.__dict__['temp']

Okay now let’s use our change method:

>>> a.change('Intervals')
>>> print(a.temp)
>>> print(A.temp)

Well now that we have used self, print(a.temp) gives us a different value from print(A.temp).

Now if we compare id(a.temp) and id(A.temp), they will be different.

Answered By: Swapnil

To protect your variable shared by other instance you need to create new instance variable each time you create an instance. When you are declaring a variable inside a class it’s class variable and shared by all instance. If you want to make it for instance wise need to use the init method to reinitialize the variable as refer to the instance

From Python Objects and Class by

__init__() function. This special function gets called whenever a new object of that class is instantiated.

This type of function is also called constructors in Object Oriented
Programming (OOP). We normally use it to initialize all the variables.

For example:

class example:
    list=[] #This is class variable shared by all instance
    def __init__(self):
        self.list = [] #This is instance variable referred to specific instance
Answered By: Projesh Bhoumik
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