What I want is this behavior:
class a:
list = []
x = a()
y = a()
x.list.append(1)
y.list.append(2)
x.list.append(3)
y.list.append(4)
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?
You want this:
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.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 Programiz.com:
For example:
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:
So what do we have here?
temp
which is a stringself.x
Pretty straight forward so far yeah? Now let's start playing around with this class. Let's initialize this class first:
Now do the following:
Well,
a.temp
worked as expected but how the hell didA.temp
work? Well it worked because temp is a class attribute. Everything in python is an object. Here A is also an object of classtype
. 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: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:
Note that
temp
attribute is listed among A class's attributes while x is listed for the instanceSo 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 writea.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 casea.__getattribute__('temp')
executes firsta.__dict__['temp']
and thena.__class__.__dict__['temp']
Okay now let's use our
change
method:Well now that we have used self,
print a.temp
gives us a different value fromprint A.temp
.Now if we compare id(a.temp) and id(A.temp), they will be different
I think the answers provided are misleading. A property defined inside a class becomes an instance property when the object is instantiated, regardless of how you define it. So copies of
a.list
are made, andx.list
andy.list
are different copies. The reason they seem to be the same is that they are both aliases to the same list. But that is a consequence of the way lists work, not of the way classes work. If you were to do the same thing with numbers instead of lists (or just using += instead of append, which would create a new list) you would see that changingx.attr
doesn't affect changingy.attr
.Defining
self.list
inside__init__
works, because the function is called twice, once for each time the object is instantiated, and so, two different lists are created.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: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:
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.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.
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.