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The += operator in python seems to be operating unexpectedly on lists. Can anyone tell me what is going on here?
class foo:
bar = []
def __init__(self,x):
self.bar += [x]
class foo2:
bar = []
def __init__(self,x):
self.bar = self.bar + [x]
f = foo(1)
g = foo(2)
print f.bar
print g.bar
f.bar += [3]
print f.bar
print g.bar
f.bar = f.bar + [4]
print f.bar
print g.bar
f = foo2(1)
g = foo2(2)
print f.bar
print g.bar
OUTPUT
[1, 2]
[1, 2]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3, 4]
[1, 2, 3]
[1]
[2]
foo += bar seems to affect every instance of the class, whereas foo = foo + bar seems to behave in the way I would expect things to behave.
The += operator is called a "compound assignment operator".
The general answer is that += tries to call the __iadd__ special method, and if that isn't available it tries to use __add__ instead. So the issue is with the difference between these special methods.
The __iadd__ special method is for an in-place addition, that is it mutates the object that it acts on. The __add__ special method returns a new object and is also used for the standard + operator.
So when the += operator is used on an object which has an __iadd__ defined the object is modified in place. Otherwise it will instead try to use the plain __add__ and return a new object.
That is why for mutable types like lists += changes the object's value, whereas for immutable types like tuples, strings and integers a new object is returned instead (a += b becomes equivalent to a = a + b).
For types that support both __iadd__ and __add__ you therefore have to be careful which one you use. a += b will call __iadd__ and mutate a, whereas a = a + b will create a new object and assign it to a. They are not the same operation!
>>> a1 = a2 = [1, 2]
>>> b1 = b2 = [1, 2]
>>> a1 += [3] # Uses __iadd__, modifies a1 in-place
>>> b1 = b1 + [3] # Uses __add__, creates new list, assigns it to b1
>>> a2
[1, 2, 3] # a1 and a2 are still the same list
>>> b2
[1, 2] # whereas only b1 was changed
For immutable types (where you don't have an __iadd__) a += b and a = a + b are equivalent. This is what lets you use += on immutable types, which might seem a strange design decision until you consider that otherwise you couldn't use += on immutable types like numbers!
For the general case, see Scott Griffith's answer. When dealing with lists like you are, though, the += operator is a shorthand for someListObject.extend(iterableObject). See the documentation of extend().
The extend function will append all elements of the parameter to the list.
When doing foo += something you're modifying the list foo in place, thus you don't change the reference that the name foo points to, but you're changing the list object directly. With foo = foo + something, you're actually creating a new list.
This example code will explain it:
>>> l = []
>>> id(l)
13043192
>>> l += [3]
>>> id(l)
13043192
>>> l = l + [3]
>>> id(l)
13059216
Note how the reference changes when you reassign the new list to l.
As bar is a class variable instead of an instance variable, modifying in place will affect all instances of that class. But when redefining self.bar, the instance will have a separate instance variable self.bar without affecting the other class instances.
The problem here is, bar is defined as a class attribute, not an instance variable.
In foo, the class attribute is modified in the init method, that's why all instances are affected.
In foo2, an instance variable is defined using the (empty) class attribute, and every instance gets its own bar.
The "correct" implementation would be:
class foo:
def __init__(self, x):
self.bar = [x]
Of course, class attributes are completely legal. In fact, you can access and modify them without creating an instance of the class like this:
class foo:
bar = []
foo.bar = [x]
There are two things involved here:
1. class attributes and instance attributes
2. difference between the operators + and += for lists
+ operator calls the __add__ method on a list. It takes all the elements from its operands and makes a new list containing those elements maintaining their order.
+= operator calls __iadd__ method on the list. It takes an iterable and appends all the elements of the iterable to the list in place. It does not create a new list object.
In class foo the statement self.bar += [x] is not an assignment statement but actually translates to
self.bar.__iadd__([x]) # modifies the class attribute
which modifies the list in place and acts like the list method extend.
In class foo2, on the contrary, the assignment statement in the init method
self.bar = self.bar + [x]
can be deconstructed as:
The instance has no attribute bar (there is a class attribute of the same name, though) so it accesses the class attribute bar and creates a new list by appending x to it. The statement translates to:
self.bar = self.bar.__add__([x]) # bar on the lhs is the class attribute
Then it creates an instance attribute bar and assigns the newly created list to it. Note that bar on the rhs of the assignment is different from the bar on the lhs.
For instances of class foo, bar is a class attribute and not instance attribute. Hence any change to the class attribute bar will be reflected for all instances.
On the contrary, each instance of the class foo2 has its own instance attribute bar which is different from the class attribute of the same name bar.
f = foo2(4)
print f.bar # accessing the instance attribute. prints [4]
print f.__class__.bar # accessing the class attribute. prints []
Hope this clears things.
Although much time has passed and many correct things were said, there is no answer which bundles both effects.
You have 2 effects:
a "special", maybe unnoticed behaviour of lists with += (as stated by Scott Griffiths)
the fact that class attributes as well as instance attributes are involved (as stated by Can Berk Büder)
In class foo, the __init__ method modifies the class attribute. It is because self.bar += [x] translates to self.bar = self.bar.__iadd__([x]). __iadd__() is for inplace modification, so it modifies the list and returns a reference to it.
Note that the instance dict is modified although this would normally not be necessary as the class dict already contains the same assignment. So this detail goes almost unnoticed - except if you do a foo.bar = [] afterwards. Here the instances's bar stays the same thanks to the said fact.
In class foo2, however, the class's bar is used, but not touched. Instead, a [x] is added to it, forming a new object, as self.bar.__add__([x]) is called here, which doesn't modify the object. The result is put into the instance dict then, giving the instance the new list as a dict, while the class's attribute stays modified.
The distinction between ... = ... + ... and ... += ... affects as well the assignments afterwards:
f = foo(1) # adds 1 to the class's bar and assigns f.bar to this as well.
g = foo(2) # adds 2 to the class's bar and assigns g.bar to this as well.
# Here, foo.bar, f.bar and g.bar refer to the same object.
print f.bar # [1, 2]
print g.bar # [1, 2]
f.bar += [3] # adds 3 to this object
print f.bar # As these still refer to the same object,
print g.bar # the output is the same.
f.bar = f.bar + [4] # Construct a new list with the values of the old ones, 4 appended.
print f.bar # Print the new one
print g.bar # Print the old one.
f = foo2(1) # Here a new list is created on every call.
g = foo2(2)
print f.bar # So these all obly have one element.
print g.bar
You can verify the identity of the objects with print id(foo), id(f), id(g) (don't forget the additional ()s if you are on Python3).
BTW: The += operator is called "augmented assignment" and generally is intended to do inplace modifications as far as possible.
The other answers would seem to pretty much have it covered, though it seems worth quoting and referring to the Augmented Assignments PEP 203:
They [the augmented assignment operators] implement the same operator
as their normal binary form, except that the operation is done
`in-place' when the left-hand side object supports it, and that the
left-hand side is only evaluated once.
...
The idea behind augmented
assignment in Python is that it isn't just an easier way to write the
common practice of storing the result of a binary operation in its
left-hand operand, but also a way for the left-hand operand in
question to know that it should operate `on itself', rather than
creating a modified copy of itself.
>>> elements=[[1],[2],[3]]
>>> subset=[]
>>> subset+=elements[0:1]
>>> subset
[[1]]
>>> elements
[[1], [2], [3]]
>>> subset[0][0]='change'
>>> elements
[['change'], [2], [3]]
>>> a=[1,2,3,4]
>>> b=a
>>> a+=[5]
>>> a,b
([1, 2, 3, 4, 5], [1, 2, 3, 4, 5])
>>> a=[1,2,3,4]
>>> b=a
>>> a=a+[5]
>>> a,b
([1, 2, 3, 4, 5], [1, 2, 3, 4])
>>> a = 89
>>> id(a)
4434330504
>>> a = 89 + 1
>>> print(a)
90
>>> id(a)
4430689552 # this is different from before!
>>> test = [1, 2, 3]
>>> id(test)
48638344L
>>> test2 = test
>>> id(test)
48638344L
>>> test2 += [4]
>>> id(test)
48638344L
>>> print(test, test2) # [1, 2, 3, 4] [1, 2, 3, 4]```
([1, 2, 3, 4], [1, 2, 3, 4])
>>> id(test2)
48638344L # ID is different here
We see that when we attempt to modify an immutable object (integer in this case), Python simply gives us a different object instead. On the other hand, we are able to make changes to an mutable object (a list) and have it remain the same object throughout.
ref : https://medium.com/#tyastropheus/tricky-python-i-memory-management-for-mutable-immutable-objects-21507d1e5b95
Also refer below url to understand the shallowcopy and deepcopy
https://www.geeksforgeeks.org/copy-python-deep-copy-shallow-copy/
listname.extend() works great for this purpose :)
The += operator in python seems to be operating unexpectedly on lists. Can anyone tell me what is going on here?
class foo:
bar = []
def __init__(self,x):
self.bar += [x]
class foo2:
bar = []
def __init__(self,x):
self.bar = self.bar + [x]
f = foo(1)
g = foo(2)
print f.bar
print g.bar
f.bar += [3]
print f.bar
print g.bar
f.bar = f.bar + [4]
print f.bar
print g.bar
f = foo2(1)
g = foo2(2)
print f.bar
print g.bar
OUTPUT
[1, 2]
[1, 2]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3, 4]
[1, 2, 3]
[1]
[2]
foo += bar seems to affect every instance of the class, whereas foo = foo + bar seems to behave in the way I would expect things to behave.
The += operator is called a "compound assignment operator".
The general answer is that += tries to call the __iadd__ special method, and if that isn't available it tries to use __add__ instead. So the issue is with the difference between these special methods.
The __iadd__ special method is for an in-place addition, that is it mutates the object that it acts on. The __add__ special method returns a new object and is also used for the standard + operator.
So when the += operator is used on an object which has an __iadd__ defined the object is modified in place. Otherwise it will instead try to use the plain __add__ and return a new object.
That is why for mutable types like lists += changes the object's value, whereas for immutable types like tuples, strings and integers a new object is returned instead (a += b becomes equivalent to a = a + b).
For types that support both __iadd__ and __add__ you therefore have to be careful which one you use. a += b will call __iadd__ and mutate a, whereas a = a + b will create a new object and assign it to a. They are not the same operation!
>>> a1 = a2 = [1, 2]
>>> b1 = b2 = [1, 2]
>>> a1 += [3] # Uses __iadd__, modifies a1 in-place
>>> b1 = b1 + [3] # Uses __add__, creates new list, assigns it to b1
>>> a2
[1, 2, 3] # a1 and a2 are still the same list
>>> b2
[1, 2] # whereas only b1 was changed
For immutable types (where you don't have an __iadd__) a += b and a = a + b are equivalent. This is what lets you use += on immutable types, which might seem a strange design decision until you consider that otherwise you couldn't use += on immutable types like numbers!
For the general case, see Scott Griffith's answer. When dealing with lists like you are, though, the += operator is a shorthand for someListObject.extend(iterableObject). See the documentation of extend().
The extend function will append all elements of the parameter to the list.
When doing foo += something you're modifying the list foo in place, thus you don't change the reference that the name foo points to, but you're changing the list object directly. With foo = foo + something, you're actually creating a new list.
This example code will explain it:
>>> l = []
>>> id(l)
13043192
>>> l += [3]
>>> id(l)
13043192
>>> l = l + [3]
>>> id(l)
13059216
Note how the reference changes when you reassign the new list to l.
As bar is a class variable instead of an instance variable, modifying in place will affect all instances of that class. But when redefining self.bar, the instance will have a separate instance variable self.bar without affecting the other class instances.
The problem here is, bar is defined as a class attribute, not an instance variable.
In foo, the class attribute is modified in the init method, that's why all instances are affected.
In foo2, an instance variable is defined using the (empty) class attribute, and every instance gets its own bar.
The "correct" implementation would be:
class foo:
def __init__(self, x):
self.bar = [x]
Of course, class attributes are completely legal. In fact, you can access and modify them without creating an instance of the class like this:
class foo:
bar = []
foo.bar = [x]
There are two things involved here:
1. class attributes and instance attributes
2. difference between the operators + and += for lists
+ operator calls the __add__ method on a list. It takes all the elements from its operands and makes a new list containing those elements maintaining their order.
+= operator calls __iadd__ method on the list. It takes an iterable and appends all the elements of the iterable to the list in place. It does not create a new list object.
In class foo the statement self.bar += [x] is not an assignment statement but actually translates to
self.bar.__iadd__([x]) # modifies the class attribute
which modifies the list in place and acts like the list method extend.
In class foo2, on the contrary, the assignment statement in the init method
self.bar = self.bar + [x]
can be deconstructed as:
The instance has no attribute bar (there is a class attribute of the same name, though) so it accesses the class attribute bar and creates a new list by appending x to it. The statement translates to:
self.bar = self.bar.__add__([x]) # bar on the lhs is the class attribute
Then it creates an instance attribute bar and assigns the newly created list to it. Note that bar on the rhs of the assignment is different from the bar on the lhs.
For instances of class foo, bar is a class attribute and not instance attribute. Hence any change to the class attribute bar will be reflected for all instances.
On the contrary, each instance of the class foo2 has its own instance attribute bar which is different from the class attribute of the same name bar.
f = foo2(4)
print f.bar # accessing the instance attribute. prints [4]
print f.__class__.bar # accessing the class attribute. prints []
Hope this clears things.
Although much time has passed and many correct things were said, there is no answer which bundles both effects.
You have 2 effects:
a "special", maybe unnoticed behaviour of lists with += (as stated by Scott Griffiths)
the fact that class attributes as well as instance attributes are involved (as stated by Can Berk Büder)
In class foo, the __init__ method modifies the class attribute. It is because self.bar += [x] translates to self.bar = self.bar.__iadd__([x]). __iadd__() is for inplace modification, so it modifies the list and returns a reference to it.
Note that the instance dict is modified although this would normally not be necessary as the class dict already contains the same assignment. So this detail goes almost unnoticed - except if you do a foo.bar = [] afterwards. Here the instances's bar stays the same thanks to the said fact.
In class foo2, however, the class's bar is used, but not touched. Instead, a [x] is added to it, forming a new object, as self.bar.__add__([x]) is called here, which doesn't modify the object. The result is put into the instance dict then, giving the instance the new list as a dict, while the class's attribute stays modified.
The distinction between ... = ... + ... and ... += ... affects as well the assignments afterwards:
f = foo(1) # adds 1 to the class's bar and assigns f.bar to this as well.
g = foo(2) # adds 2 to the class's bar and assigns g.bar to this as well.
# Here, foo.bar, f.bar and g.bar refer to the same object.
print f.bar # [1, 2]
print g.bar # [1, 2]
f.bar += [3] # adds 3 to this object
print f.bar # As these still refer to the same object,
print g.bar # the output is the same.
f.bar = f.bar + [4] # Construct a new list with the values of the old ones, 4 appended.
print f.bar # Print the new one
print g.bar # Print the old one.
f = foo2(1) # Here a new list is created on every call.
g = foo2(2)
print f.bar # So these all obly have one element.
print g.bar
You can verify the identity of the objects with print id(foo), id(f), id(g) (don't forget the additional ()s if you are on Python3).
BTW: The += operator is called "augmented assignment" and generally is intended to do inplace modifications as far as possible.
The other answers would seem to pretty much have it covered, though it seems worth quoting and referring to the Augmented Assignments PEP 203:
They [the augmented assignment operators] implement the same operator
as their normal binary form, except that the operation is done
`in-place' when the left-hand side object supports it, and that the
left-hand side is only evaluated once.
...
The idea behind augmented
assignment in Python is that it isn't just an easier way to write the
common practice of storing the result of a binary operation in its
left-hand operand, but also a way for the left-hand operand in
question to know that it should operate `on itself', rather than
creating a modified copy of itself.
>>> elements=[[1],[2],[3]]
>>> subset=[]
>>> subset+=elements[0:1]
>>> subset
[[1]]
>>> elements
[[1], [2], [3]]
>>> subset[0][0]='change'
>>> elements
[['change'], [2], [3]]
>>> a=[1,2,3,4]
>>> b=a
>>> a+=[5]
>>> a,b
([1, 2, 3, 4, 5], [1, 2, 3, 4, 5])
>>> a=[1,2,3,4]
>>> b=a
>>> a=a+[5]
>>> a,b
([1, 2, 3, 4, 5], [1, 2, 3, 4])
>>> a = 89
>>> id(a)
4434330504
>>> a = 89 + 1
>>> print(a)
90
>>> id(a)
4430689552 # this is different from before!
>>> test = [1, 2, 3]
>>> id(test)
48638344L
>>> test2 = test
>>> id(test)
48638344L
>>> test2 += [4]
>>> id(test)
48638344L
>>> print(test, test2) # [1, 2, 3, 4] [1, 2, 3, 4]```
([1, 2, 3, 4], [1, 2, 3, 4])
>>> id(test2)
48638344L # ID is different here
We see that when we attempt to modify an immutable object (integer in this case), Python simply gives us a different object instead. On the other hand, we are able to make changes to an mutable object (a list) and have it remain the same object throughout.
ref : https://medium.com/#tyastropheus/tricky-python-i-memory-management-for-mutable-immutable-objects-21507d1e5b95
Also refer below url to understand the shallowcopy and deepcopy
https://www.geeksforgeeks.org/copy-python-deep-copy-shallow-copy/
listname.extend() works great for this purpose :)
Let's suppose I have 2 classes in different scenario.
Scenario 1
class MyClass():
temp = 5
Scenario 2
class MyClass():
temp = 5
def myfunc(self):
print self.temp
Now when will variable temp will be treated as a class variable and instance variable. I am confused because in both the scenarios I am able to access the value of variable temp using both.
Object.Temp (behaving as instance variable)
ClassName.Temp (behaving as class variable)
I believe similar questions have been asked before but it will be a great help if someone can explain this in context of my question.
Class variables are shared between all instances of a class. With immutable types (like int, str, ...) you won't note much of a difference. But consider this:
class MyClass():
temp = []
def myfunc(self, val):
self.temp.append(val)
print self.temp
instance1 = MyClass()
instance1.myfunc(1) # [1]
instance2 = MyClass()
instance2.myfunc(2) # [1, 2]
In this case both instances share the same list, that is if the instance doesn't have a temp member itself, then that of the class is used.
So if you further do:
MyClass.temp.append(3)
print instance1.temp # [1, 2, 3]
instance1.temp = []
print instance1.temp # [] uses the instances temp
print instance2.temp # [1, 2, 3]
del instance1.temp
print instance1.temp # [1, 2, 3] uses the class' temp again
Basically, MyClass.temp is always a class variable. Getting obj.temp returns the class variable, until you try to set obj.temp, which creates a member variable that masks the class variable. I hope this helps:
>>> class MyClass(object):
... temp = 5
...
>>> a = MyClass()
>>> b = MyClass()
>>> a.temp
5
>>> b.temp
5
>>> b.temp = 6
>>> a.temp
5
>>> MyClass.temp = 7
>>> a.temp
7
>>> b.temp
6
>>> a.__dict__
{}
>>> b.__dict__
{'temp': 6}
>>> MyClass.__dict__
{..., 'temp': 7}
Edit: As mata says, calling methods (such as append()) on obj.temp does not count as "setting" it.
temp is a class variable. When you access the variable it is searched through the layers of inheritance so since it is not found in the instance itself it checks the class(next layer up) and finds it there.
Can someone explain why Python does the following?
>>> class Foo(object):
... bar = []
...
>>> a = Foo()
>>> b = Foo()
>>> a.bar.append(1)
>>> b.bar
[1]
>>> a.bar = 1
>>> a.bar
1
>>> b.bar
[1]
>>> a.bar = []
>>> a.bar
[]
>>> b.bar
[1]
>>> del a.bar
>>> a.bar
[1]
It's rather confusing!
This is because the way you have written it, bar is a class variable rather than an instance variable.
To define an instance variable, bind it in the constructor:
class Foo(object):
def __init__(self):
self.bar = []
Note that it now belongs to a single instance of Foo (self) rather than the Foo class, and you will see the results you expect when you assign to it.
As others have said the code as written creates a class variable rather than an instance variable. You need to assign in __init__ to create an instance variable.
Hopefully this annotated copy of your code is helpful in explaining what's going on at each stage:
>>> class Foo(object):
... bar = [] # defines a class variable on Foo (shared by all instances)
...
>>> a = Foo()
>>> b = Foo()
>>> a.bar.append(1) # appends the value 1 to the previously empty list Foo.bar
>>> b.bar # returns the value of the class variable Foo.bar
[1]
>>> a.bar = 1 # binds 1 to the instance variable a.bar, masking the access
>>> a.bar # you previously had to the class variable through a.bar
1
>>> b.bar # b doesn't have an instance variable 'bar' so this still
[1] # returns the class variable
>>> a.bar = [] # bind a's instance variable to to an empty list
>>> a.bar
[]
>>> b.bar # b doesn't have an instance variable 'bar' so this still
[1] # returns the class variable
>>> del a.bar # unbinds a's instance variable unmasking the class variable
>>> a.bar # so a.bar now returns the list with 1 in it.
[1]
Also, printing out the value of Foo.bar (the class variable accessed via the class rather than via an instance) after each of your statements might help clarify what is going on.
When you declare an element in the class like that it is shared by all instances of the class. To make a proper class member that belongs to each instance, separately, create it in __init__ like the following:
class Foo(object):
def __init__(self):
self.bar = []
In the beginning, bar is a class variable and it is shared between a and b, both a.bar and b.bar refer to the same object.
When you assign a new value to a.bar, this does not overwrite the class variable, it adds a new instance variable to the a object, hiding the class variable when you access a.bar. If you delete a.bar (the instance variable), then a.bar resolves again to the class variable.
b.bar on the other hand always refers to the class variable, it's not influenced by the additional bar on the a object or any values assigned to that.
To set the class variable you can access it through the class itself:
Foo.bar = 1
>>> class Foo(object):
... bar = []
...
bar is a shared class variable, not an instance variable. I believe that deals with most of your confusion. To make it a instance var, define it in class's __init__ per the other answers.
>>> a = Foo()
>>> b = Foo()
>>> a.bar.append(1)
>>> b.bar
[1]
This is the proof of that.
>>> a.bar = 1
>>> a.bar
1
>>> b.bar
[1]
Now you've redefined a.bar as a instance variable. That's what happens when you define variables externally by default.
>>> a.bar = []
>>> a.bar
[]
>>> b.bar
[1]
>>> del a.bar
>>> a.bar
[1]
Same again. b.bar is still the shared class variable.
On a related note, you should be aware of this pitfall that you might see sometime soon:
class A:
def __init__(self, mylist = []):
self.mylist = mylist
a = A()
a2 = A()
a.mylist.append(3)
print b.mylist #prints [3] ???
This confuses a lot of folks and has to do with how the code is interpreted. Python actually interprets the function headings first, so it evaluates __init__(self, mylist = []) and stores a reference to that list as the default parameter. That means that all instances of A will (unless provided their own list) reference the original list. The correct code for doing such a thing would be
class A:
def __init__(self, mylist=None):
if mylist:
self.mylist = mylist
else:
self.mylist = []
or if you want a shorter expression you can use the ternary syntax:
self.mylist = mylist if mylist else []
The += operator in python seems to be operating unexpectedly on lists. Can anyone tell me what is going on here?
class foo:
bar = []
def __init__(self,x):
self.bar += [x]
class foo2:
bar = []
def __init__(self,x):
self.bar = self.bar + [x]
f = foo(1)
g = foo(2)
print f.bar
print g.bar
f.bar += [3]
print f.bar
print g.bar
f.bar = f.bar + [4]
print f.bar
print g.bar
f = foo2(1)
g = foo2(2)
print f.bar
print g.bar
OUTPUT
[1, 2]
[1, 2]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3, 4]
[1, 2, 3]
[1]
[2]
foo += bar seems to affect every instance of the class, whereas foo = foo + bar seems to behave in the way I would expect things to behave.
The += operator is called a "compound assignment operator".
The general answer is that += tries to call the __iadd__ special method, and if that isn't available it tries to use __add__ instead. So the issue is with the difference between these special methods.
The __iadd__ special method is for an in-place addition, that is it mutates the object that it acts on. The __add__ special method returns a new object and is also used for the standard + operator.
So when the += operator is used on an object which has an __iadd__ defined the object is modified in place. Otherwise it will instead try to use the plain __add__ and return a new object.
That is why for mutable types like lists += changes the object's value, whereas for immutable types like tuples, strings and integers a new object is returned instead (a += b becomes equivalent to a = a + b).
For types that support both __iadd__ and __add__ you therefore have to be careful which one you use. a += b will call __iadd__ and mutate a, whereas a = a + b will create a new object and assign it to a. They are not the same operation!
>>> a1 = a2 = [1, 2]
>>> b1 = b2 = [1, 2]
>>> a1 += [3] # Uses __iadd__, modifies a1 in-place
>>> b1 = b1 + [3] # Uses __add__, creates new list, assigns it to b1
>>> a2
[1, 2, 3] # a1 and a2 are still the same list
>>> b2
[1, 2] # whereas only b1 was changed
For immutable types (where you don't have an __iadd__) a += b and a = a + b are equivalent. This is what lets you use += on immutable types, which might seem a strange design decision until you consider that otherwise you couldn't use += on immutable types like numbers!
For the general case, see Scott Griffith's answer. When dealing with lists like you are, though, the += operator is a shorthand for someListObject.extend(iterableObject). See the documentation of extend().
The extend function will append all elements of the parameter to the list.
When doing foo += something you're modifying the list foo in place, thus you don't change the reference that the name foo points to, but you're changing the list object directly. With foo = foo + something, you're actually creating a new list.
This example code will explain it:
>>> l = []
>>> id(l)
13043192
>>> l += [3]
>>> id(l)
13043192
>>> l = l + [3]
>>> id(l)
13059216
Note how the reference changes when you reassign the new list to l.
As bar is a class variable instead of an instance variable, modifying in place will affect all instances of that class. But when redefining self.bar, the instance will have a separate instance variable self.bar without affecting the other class instances.
The problem here is, bar is defined as a class attribute, not an instance variable.
In foo, the class attribute is modified in the init method, that's why all instances are affected.
In foo2, an instance variable is defined using the (empty) class attribute, and every instance gets its own bar.
The "correct" implementation would be:
class foo:
def __init__(self, x):
self.bar = [x]
Of course, class attributes are completely legal. In fact, you can access and modify them without creating an instance of the class like this:
class foo:
bar = []
foo.bar = [x]
There are two things involved here:
1. class attributes and instance attributes
2. difference between the operators + and += for lists
+ operator calls the __add__ method on a list. It takes all the elements from its operands and makes a new list containing those elements maintaining their order.
+= operator calls __iadd__ method on the list. It takes an iterable and appends all the elements of the iterable to the list in place. It does not create a new list object.
In class foo the statement self.bar += [x] is not an assignment statement but actually translates to
self.bar.__iadd__([x]) # modifies the class attribute
which modifies the list in place and acts like the list method extend.
In class foo2, on the contrary, the assignment statement in the init method
self.bar = self.bar + [x]
can be deconstructed as:
The instance has no attribute bar (there is a class attribute of the same name, though) so it accesses the class attribute bar and creates a new list by appending x to it. The statement translates to:
self.bar = self.bar.__add__([x]) # bar on the lhs is the class attribute
Then it creates an instance attribute bar and assigns the newly created list to it. Note that bar on the rhs of the assignment is different from the bar on the lhs.
For instances of class foo, bar is a class attribute and not instance attribute. Hence any change to the class attribute bar will be reflected for all instances.
On the contrary, each instance of the class foo2 has its own instance attribute bar which is different from the class attribute of the same name bar.
f = foo2(4)
print f.bar # accessing the instance attribute. prints [4]
print f.__class__.bar # accessing the class attribute. prints []
Hope this clears things.
Although much time has passed and many correct things were said, there is no answer which bundles both effects.
You have 2 effects:
a "special", maybe unnoticed behaviour of lists with += (as stated by Scott Griffiths)
the fact that class attributes as well as instance attributes are involved (as stated by Can Berk Büder)
In class foo, the __init__ method modifies the class attribute. It is because self.bar += [x] translates to self.bar = self.bar.__iadd__([x]). __iadd__() is for inplace modification, so it modifies the list and returns a reference to it.
Note that the instance dict is modified although this would normally not be necessary as the class dict already contains the same assignment. So this detail goes almost unnoticed - except if you do a foo.bar = [] afterwards. Here the instances's bar stays the same thanks to the said fact.
In class foo2, however, the class's bar is used, but not touched. Instead, a [x] is added to it, forming a new object, as self.bar.__add__([x]) is called here, which doesn't modify the object. The result is put into the instance dict then, giving the instance the new list as a dict, while the class's attribute stays modified.
The distinction between ... = ... + ... and ... += ... affects as well the assignments afterwards:
f = foo(1) # adds 1 to the class's bar and assigns f.bar to this as well.
g = foo(2) # adds 2 to the class's bar and assigns g.bar to this as well.
# Here, foo.bar, f.bar and g.bar refer to the same object.
print f.bar # [1, 2]
print g.bar # [1, 2]
f.bar += [3] # adds 3 to this object
print f.bar # As these still refer to the same object,
print g.bar # the output is the same.
f.bar = f.bar + [4] # Construct a new list with the values of the old ones, 4 appended.
print f.bar # Print the new one
print g.bar # Print the old one.
f = foo2(1) # Here a new list is created on every call.
g = foo2(2)
print f.bar # So these all obly have one element.
print g.bar
You can verify the identity of the objects with print id(foo), id(f), id(g) (don't forget the additional ()s if you are on Python3).
BTW: The += operator is called "augmented assignment" and generally is intended to do inplace modifications as far as possible.
The other answers would seem to pretty much have it covered, though it seems worth quoting and referring to the Augmented Assignments PEP 203:
They [the augmented assignment operators] implement the same operator
as their normal binary form, except that the operation is done
`in-place' when the left-hand side object supports it, and that the
left-hand side is only evaluated once.
...
The idea behind augmented
assignment in Python is that it isn't just an easier way to write the
common practice of storing the result of a binary operation in its
left-hand operand, but also a way for the left-hand operand in
question to know that it should operate `on itself', rather than
creating a modified copy of itself.
>>> elements=[[1],[2],[3]]
>>> subset=[]
>>> subset+=elements[0:1]
>>> subset
[[1]]
>>> elements
[[1], [2], [3]]
>>> subset[0][0]='change'
>>> elements
[['change'], [2], [3]]
>>> a=[1,2,3,4]
>>> b=a
>>> a+=[5]
>>> a,b
([1, 2, 3, 4, 5], [1, 2, 3, 4, 5])
>>> a=[1,2,3,4]
>>> b=a
>>> a=a+[5]
>>> a,b
([1, 2, 3, 4, 5], [1, 2, 3, 4])
>>> a = 89
>>> id(a)
4434330504
>>> a = 89 + 1
>>> print(a)
90
>>> id(a)
4430689552 # this is different from before!
>>> test = [1, 2, 3]
>>> id(test)
48638344L
>>> test2 = test
>>> id(test)
48638344L
>>> test2 += [4]
>>> id(test)
48638344L
>>> print(test, test2) # [1, 2, 3, 4] [1, 2, 3, 4]```
([1, 2, 3, 4], [1, 2, 3, 4])
>>> id(test2)
48638344L # ID is different here
We see that when we attempt to modify an immutable object (integer in this case), Python simply gives us a different object instead. On the other hand, we are able to make changes to an mutable object (a list) and have it remain the same object throughout.
ref : https://medium.com/#tyastropheus/tricky-python-i-memory-management-for-mutable-immutable-objects-21507d1e5b95
Also refer below url to understand the shallowcopy and deepcopy
https://www.geeksforgeeks.org/copy-python-deep-copy-shallow-copy/
listname.extend() works great for this purpose :)