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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 :)
When I have a function like
def foo(A):
tmp=A
tmp=tmp+1
*rest of the function*
return
or
def foo(A):
global tmp
tmp=A
tmp=tmp+1
*rest of the function*
return
both to "A" and "tmp" 1 is added instead of only to "tmp". What am I doing wrong and how can I fix it?
As nneonneo's comment under Blake's answer (which describes a significant part of the problem) states, the code in the original post doesn't actually do what he states. For example:
def foo(A):
tmp = A
tmp = tmp + 1
return (A, tmp)
foo(3)
returns (3,4), meaning A has been left unchanged.
Where this would not be true is where A is a mutable type. Mutable types include lists, dictionaries, and derived types of those, while integers, floats, and tuples are not mutable.
For example:
def foo(A):
tmp = A
tmp[0] = tmp[0] + 1
return (A, tmp)
foo([1, 2])
which returns ([2, 2], [2, 2]), and in this case A has changed.
That foo changes the value of A in the list case but not the integer case is because lists are mutable and integers are not. Assigning A to tmp when A is a mutable type assigns a reference to the mutable object, and changing one of its elements (as in tmp[0] = tmp[0] + 1) doesn't make a new object.
If you do not want your function to have this side-effect-like behavior for a list, for example, a common Python idiom is to use slice notation to duplicate the list. This will make a new list object when you assign it to tmp that is a copy of the list object in A:
def foo(A):
tmp = A[:]
# this slice makes a new list, a copy of A
tmp[0] = tmp[0] + 1
return (A, tmp)
foo([1, 2])
This returns ([1, 2], [2, 2]), so A is unchanged and tmp is changed.
There are other ways to copy lists or other mutable objects which are subtly different from each other. How to clone or copy a list? has a great description of your choices.
That's because python method parameters are pass by reference, not pass by value. You're essentially modifying the same place in memory, that two different variables point to.
>>> def foo(a):
tmp = a
print(tmp, a, id(a), id(tmp))
>>> foo(5)
5 5 505910928 505910928
>>> b = 5
>>> foo(b)
5 5 505910928 505910928
>>> id(b)
505910928
And with the global example:
>>> def foo(a):
global tmp
a = tmp
print(a, tmp, id(a), id(tmp))
>>> foo(5)
7 7 505910960 505910960
>>> foo('s')
7 7 505910960 505910960
>>> tmp
7
>>> tmp = 6
>>> foo('a')
6 6 505910944 505910944
I am just learning python and I am going though the tutorials on https://developers.google.com/edu/python/strings
Under the String Slices section
s[:] is 'Hello' -- omitting both always gives us a copy of the whole
thing (this is the pythonic way to copy a sequence like a string or
list)
Out of curiosity why wouldn't you just use an = operator?
s = 'hello';
bar = s[:]
foo = s
As far as I can tell both bar and foo have the same value.
= makes a reference, by using [:] you create a copy. For strings, which are immutable, this doesn't really matter, but for lists etc. it is crucial.
>>> s = 'hello'
>>> t1 = s
>>> t2 = s[:]
>>> print s, t1, t2
hello hello hello
>>> s = 'good bye'
>>> print s, t1, t2
good bye hello hello
but:
>>> li1 = [1,2]
>>> li = [1,2]
>>> li1 = li
>>> li2 = li[:]
>>> print li, li1, li2
[1, 2] [1, 2] [1, 2]
>>> li[0] = 0
>>> print li, li1, li2
[0, 2] [0, 2] [1, 2]
So why use it when dealing with strings? The built-in strings are immutable, but whenever you write a library function expecting a string, a user might give you something that "looks like a string" and "behaves like a string", but is a custom type. This type might be mutable, so it's better to take care of that.
Such a type might look like:
class MutableString(object):
def __init__(self, s):
self._characters = [c for c in s]
def __str__(self):
return "".join(self._characters)
def __repr__(self):
return "MutableString(\"%s\")" % str(self)
def __getattr__(self, name):
return str(self).__getattribute__(name)
def __len__(self):
return len(self._characters)
def __getitem__(self, index):
return self._characters[index]
def __setitem__(self, index, value):
self._characters[index] = value
def __getslice__(self, start, end=-1, stride=1):
return str(self)[start:end:stride]
if __name__ == "__main__":
m = MutableString("Hello")
print m
print len(m)
print m.find("o")
print m.find("x")
print m.replace("e", "a") #translate to german ;-)
print m
print m[3]
m[1] = "a"
print m
print m[:]
copy1 = m
copy2 = m[:]
print m, copy1, copy2
m[1] = "X"
print m, copy1, copy2
Disclaimer: This is just a sample to show how it could work and to motivate the use of [:]. It is untested, incomplete and probably horribly performant
They have the same value, but there is a fundamental difference when dealing with mutable objects.
Say foo = [1, 2, 3]. You assign bar = foo, and baz = foo[:]. Now let's say you want to change bar - bar.append(4). You check the value of foo, and...
print foo
# [1, 2, 3, 4]
Now where did that extra 4 come from? It's because you assigned bar to the identity of foo, so when you change one you change the other. You change baz - baz.append(5), but nothing has happened to the other two - that's because you assigned a copy of foo to baz.
Note however that because strings are immutable, it doesn't matter.
If you have a list the result is different:
l = [1,2,3]
l1 = l
l2 = l[:]
l2 is a copy of l (different object) while l1 is an alias of l which means that l1[0]=7 will modify also l, while l2[1]=7 will not modify l.
While referencing an object and referencing the object's copy doesn't differ for an immutable object like string, they do for mutable objects (and mutable methods), for instance list.
Same thing on mutable objects:
a = [1,2,3,4]
b = a
c = a[:]
a[0] = -1
print a # will print [1,2,3,4]
print b # will print [-1,2,3,4]
print c # will print [1,2,3,4]
A visualization on pythontutor of the above example - http://goo.gl/Aswnl.
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 :)