I have a vector class and I defined the __mul__ method to multiply a vector by a number.
Here is the __mul__ method :
def __mul__(self, other):
x = self.x * other
y = self.y * other
new = Vector()
new.set_pos((x, y))
return new
My problem is that I don't know which is which between the number and the vector.
If self is the number, self.x raises an error. (I'm maybe mistaking on this point : Is "other" always a number ?)
So I found here : Python: multiplication override
that I could do :
__rmul__ = __mul__
but how can I do that in a class definition ?
Something like :
def __rmul__ = __mul__
self will never be the number in __mul__() because the object the method is attached to is not the number, it's the vector, and by definition it's the multiplicand.
other will be a number if your object is being multiplied by a number. Or it could be something else, such as another vector, which you could test for and handle.
When your object is the multiplier, __rmul__() is called if the multiplicand doesn't know how to handle the operation.
To handle the case in which __mul__ and __rmul__ should be the same method, because the operation is commutative, you can just do the assignment in your class definition.
class Vector(object):
def __mul__(self, other):
pass
__rmul__ = __mul__
Simply list it as an attribute:
__rmul__ = __mul__
This is the same way you'd create an alias of a function in a module; creating an alias of a method within a class body works the same.
The point is that in Python, you can tell objects how to multiply themselves by things. That means that
a * b
could either mean "tell a to multiply itself by b" or "tell b to multiply itself by a". In code, that translates to
a.__mul__(b)
or
b.__rmul__(a)
Related
I have a vector class:
class Vector:
def __init__(self, x, y):
self.x, self.y = x, y
def __str__(self):
return '(%s,%s)' % (self.x, self.y)
def __add__(self, n):
if isinstance(n, (int, long, float)):
return Vector(self.x+n, self.y+n)
elif isinstance(n, Vector):
return Vector(self.x+n.x, self.y+n.y)
which works fine, i.e. I can write:
a = Vector(1,2)
print(a + 1) # prints (2,3)
However if the order of operation is reversed, then it fails:
a = Vector(1,2)
print(1 + a) # raises TypeError: unsupported operand type(s)
# for +: 'int' and 'instance'
I understand the error: the addition of an int object to an Vector object is undefined because I haven't defined it in the int class. Is there a way to work around this without defining it in the int (or parent of int) class?
You need to also define __radd__
Some operations do not necessarily evaluate like this a + b == b + a and that's why Python defines the add and radd methods.
Explaining myself better: it supports the fact that "int" does not define a + operation with class Vector instances as part of the operation. Therefore vector + 1 is not the same as 1 + vector.
When Python tries to see what the 1.__add__ method can do, an exception is raised. And Python goes and looks for Vector.__radd__ operation to try to complete it.
In the OP's case the evaluation is true and suffices with __radd__ = __add__
class Vector(object):
def __init__(self, x, y):
self.x, self.y = x, y
def __str__(self):
return '(%s,%s)' % (self.x, self.y)
def __add__(self, n):
if isinstance(n, (int, long, float)):
return Vector(self.x+n, self.y+n)
elif isinstance(n, Vector):
return Vector(self.x+n.x, self.y+n.y)
__radd__ = __add__
a = Vector(1, 2)
print(1 + a)
Which outputs:
(2,3)
The same applies to all number-like operations.
When you say x + y, Python calls x.__add__(y). If x does not implement __add__ (or that method returns NotImplemented), Python tries to call y.__radd__(x) as a fallback.
Thus all you have to do is to define the __radd__() method in your Vector class and 1 + y will work as you would expect.
Note: you would have to do similar for other operations too, e.g. implement __mul__() and __rmul__() pair, etc.
You might also want to look at this question, it explains the same principle in more details.
Update:
Depending on your use case, you might also want to implement the __iadd__() method (and its cousins) to override the += operator.
For example, if you say y += 1 (y being an instance of Vector here), you might want to modify the y instance itself, and not return a new Vector instance as a result, which what your __add__() method currently does.
I want to overload the operator # in python for a class I have written. I know how to do operator overloading in general (i.e. by defining __add__ and __radd__ to overload +) but I could not find a way to overload #.
Why I know, that # can be overloaded: for numpy arrays, A#B gives the matrix product of A and B, while A*B gives the Hadamard (element-wise) product.
The methods you need to overload are the __matmul__ and __rmatmul__ methods. E.g. if you want to add the inner product to lists:
class Vector(list):
def __matmul__(self, other):
return sum(x * y for x, y in zip(self, other))
def __rmatmul__(self, other):
return self.__matmul__(other)
Consider the following example of a 'wrapper' class to represent vectors:
class Vector:
def __init__(self, value):
self._vals = value.copy()
def __add__(self, other):
if isinstance(other, list):
result = [x+y for (x, y) in zip(self._vals, other)]
elif isinstance(other, Vector):
result = [x+y for (x, y) in zip(self._vals, other._vals)]
else:
# assume other is scalar
result = [x+other for x in self._vals]
return Vector(result)
def __str__(self):
return str(self._vals)
The __add__ method takes care of adding two vectors as well as adding a vector with a scalar. However, the second case is not complete as the following examples show:
>>> a = Vector([1.2, 3, 4])
>>> print(a)
[1.2, 3, 4]
>>> print(a+a)
[2.4, 6, 8]
>>> print(a+5)
[6.2, 8, 9]
>>> print(5+a)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'int' and 'Vector'
To my understanding the reason is that the overloaded operator only tells Python what to do when it sees a + x where a is an instance of Vector, but there is no indication of what to do for x + a (with a an instance of Vector and x a scalar).
How one should overload the operators in such circumstances to cover all cases (i.e., to support the case that self is not an instance of Vector but other is)?
Ok. I guess I found the answer: one has to overload __radd__ operator as well:
class Vector:
def __init__(self, value):
self._vals = value.copy()
def __add__(self, other):
if isinstance(other, list):
result = [x+y for (x, y) in zip(self._vals, other)]
elif isinstance(other, Vector):
result = [x+y for (x, y) in zip(self._vals, other._vals)]
else:
# assume other is scalar
result = [x+other for x in self._vals]
return Vector(result)
def __radd__(self, other):
return self + other
def __str__(self):
return str(self._vals)
Although to me this looks a bit redundant. (Why Python does not use the commutativity of addition by default, assuming __radd__(self, other) always returns self + other? Of course for special cases the user can override __radd__.)
You could define a Scalar class that has int as its base class.
Then override __add__ to do what you want.
class Scalar(int):
def __add__(self):
# do stuff
You already figured out you need to implement __radd__. This is an answer as to why this is so, and why you need to do this in addition to implementing __add__, as a Both quotes are taken from Python Docs (Data Model - 3.3.8 Emulating numeric types), starting with the obvious:
These methods are called to implement the binary arithmetic operations (+, -, *, #, /, //, %, divmod(), pow(), **, <<, >>, &, ^, |). For instance, to evaluate the expression x + y, where x is an instance of a class that has an __add__() method, x.__add__(y) is called.
So order determines which object's implementation of __add__ is called. When the method doesn't support the operation with the passed argument NotImplemented should be returned. That's where the so-called "reflected methods" come into play:
These functions are only called if the left operand does not support the corresponding operation and the operands are of different types. For instance, to evaluate the expression x - y, where y is an instance of a class that has an __rsub__() method, y.__rsub__(x) is called if x.__sub__(y) returns NotImplemented [sic].
Now, why wouldn't __radd__(self, other) just fall back to __add__(self, other)? While ring addition is always commutative (see this and this math.stackexchange answers), you could have algebraic structures that are do not satisfy this assumption (e.g., near-rings). But my guess as a non-mathematician would be that it's just desirable to have a consistent data model across different numerical methods. While addition might be commonly commutative, multiplication is less so. (Think matrices and vectors! Although, admittedly this is not the best example, given __matmul__). I also prefer to see there being no exceptions, especially if I had to read about rings, etc. in a language documentation.
I am trying to understand how operator overriding works for two operands of a custom class.
For instance, suppose I have the following:
class Adder:
def __init__(self, value=1):
self.data = value
def __add__(self,other):
print('using __add__()')
return self.data + other
def __radd__(self,other):
print('using __radd__()')
return other + self.data
I initialize the following variables:
x = Adder(5)
y = Adder(4)
And then proceed to do the following operations:
1 + x
using __radd__()
Out[108]: 6
x + 2
using __add__()
Out[109]: 7
The two operations above seem straigtforward. If a member of my custom class is to the right of the "+" in the addition expression, then __radd__ is used. If it is on the left, then __add__ is used. This works for expressions when one operand is of the Adder type and another one is something else.
When I do this, however, I get the following result:
x + y
using __add__()
using __radd__()
Out[110]: 9
As you can see, if both operands are of the custom class, then both __add__ and __radd__ are called.
My question is how does Python unravel this situation and how is it able to call both the right-hand-addition function, as well as the left-hand-addition function.
It's because inside your methods you add the data to other. This is itself an instance of Adder. So the logic goes:
call __add__ on x;
add x.data (an int) to y (an Adder instance)
ah, right-hand operand is an instance with a __radd__ method, so
call __radd__ on y;
add int to y.data (another int).
Usually you would check to see if other was an instance of your class, and if so add other.data rather than just other.
That's the because the implementation of your __add__ and __radd__ method do not give any special treatment to the instances of the Adder class. Therefore, each __add__ call leads to an integer plus Adder instance operation which further requires __radd__ due to the Adder instance on the right side.
You can resolve this by doing:
def __add__(self, other):
print('using __add__()')
if isinstance(other, Adder):
other = other.data
return self.data + other
def __radd__(self, other):
print('using __radd__()')
return self.__add__(other)
I have a class called MyData that has __mul__ and __rmul__ defined (along with all the other arithmetic operators). Whenever these methods are used, it should always return a value of type MyData. However, I discovered that a * myDataObj is not the same as myDataObj * a, depending on the type of a. Specifically, if a was an int, it worked fine, but if a was a float then the first configuration return an array (my object has a numpy array as a member, and MyData.__getitem__ returns slices of that array) and the second configuration returns the proper value of type MyData.
Is there any way to determine the calling order of the operator in an expression like this?
Is there any way to determine the calling order of the operator in an expression like this?
First, the exact rules are described in the Data model section of the language reference, specifically the "Emulating numeric types" subsection.
The __rfoo__ methods are described as follows:
These methods are called to implement the binary arithmetic operations (+, -, *, /, %, divmod(), pow(), **, <<, >>, &, ^, |) with reflected (swapped) operands. These functions are only called if the left operand does not support the corresponding operation and the operands are of different types. [2] For instance, to evaluate the expression x - y, where y is an instance of a class that has an __rsub__() method, y.__rsub__(x) is called if x.__sub__(y) returns NotImplemented.
Note that ternary pow() will not try calling __rpow__() (the coercion rules would become too complicated).
Note If the right operand’s type is a subclass of the left operand’s type and that subclass provides the reflected method for the operation, this method will be called before the left operand’s non-reflected method. This behavior allows subclasses to override their ancestors’ operations.
Putting this into Pythonesque pseudocode, x * y is evaluated something like this:
if type(y) is type(x):
return x.__mul__(y)
elif type(y) is a subclass of type(x):
try y.__rmul__(x)
otherwise x.__mul__(y)
else:
try x.__mul__(y)
otherwise y.__rmul__(x)
Of course you can also determine the calling order dynamically by creating separate types whose methods just print their names and testing them:
class Base(object):
def __mul__(self, lhs): print('Base.mul')
def __rmul__(self, rhs): print('Base.rmul')
class Derived(Base):
def __mul__(self, lhs): print('Derived.mul')
def __rmul__(self, rhs): print('Derived.rmul')
class Unrelated(object):
def __mul__(self, lhs): print('Unrelated.mul')
def __rmul__(self, rhs): print('Unrelated.rmul')
print('Base * Base: ', end='')
Base() * Base()
for x, y in itertools.permutations((Base, Derived, Unrelated), 2):
print('{} * {}: '.format(x.__name__, y.__name__), end='')
x() * y()
What about with built in types as well?
Exactly the same rules. Since Base is not a subclass of either int or float, and neither int nor float knows how to multiply by it, they'll both call Base.__rmul__. And so will any other unrelated type you throw at it:
>>> Base() * 2
Base.mul
>>> 2 * Base()
Base.rmul
>>> Base() * 2.5
Base.mul
>>> 2.5 * Base()
Base.rmul
>>> 'sdfsdfsdfds' * Base()
Base.rmul
>>> (lambda: 23) * Base()
Base.rmul
My problem is that I'm getting different results from 1.5 * myObj and myObj * 1.5
There are a number of reasons for that:
Your __mul__ and __rmul__ code don't do the same thing.
You inherited from float.
You inherited from some builtin or extension type that handles float multiplication at the C-API level and isn't designed to allow overrides in subclasses.
You created a classic class instead of a new-style class.
You made a typo in one of the names.
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