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+= 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 +=  print f.bar print g.bar f.bar = f.bar +  print f.bar print g.bar f = foo2(1) g = foo2(2) print f.bar print g.bar
[1, 2] [1, 2] [1, 2, 3] [1, 2, 3] [1, 2, 3, 4] [1, 2, 3]  
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.
+= 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.
__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
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
__add__ you therefore have to be careful which one you use.
a += b will call
__iadd__ and mutate
a = a + b will create a new object and assign it to
a. They are not the same operation!
1, 2] b1 = b2 = [1, 2] a1 +=  # Uses __iadd__, modifies a1 in-place b1 = b1 +  # 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 changeda1 = a2 = [
For immutable types (where you don’t have an
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().
extend function will append all elements of the parameter to the list.
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:
id(l) 13043192 l +=  id(l) 13043192 l = l +  id(l) 13059216l = 
Note how the reference changes when you reassign the new list to
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.
foo, the class attribute is modified in the
init method, that’s why all instances are affected.
foo2, an instance variable is defined using the (empty) class attribute, and every instance gets its own
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.
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
foo2, on the contrary, the assignment statement in the
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
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
f = foo2(4) print f.bar # accessing the instance attribute. prints  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)
__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.
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 +=  # 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 +  # 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).
+= 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.
1],,] subset= subset+=elements[0:1] subset [] elements [, , ] subset='change' elements [['change'], , ] a=[1,2,3,4] b=a a+= a,b ([1, 2, 3, 4, 5], [1, 2, 3, 4, 5]) a=[1,2,3,4] b=a a=a+ a,b ([1, 2, 3, 4, 5], [1, 2, 3, 4])elements=[[
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 +=  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 herea =
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.
Also refer below url to understand the shallowcopy and deepcopy