I have a class called Time, and I need to implement a Frequency class. How can I implement dividing ints or floats by an instance of Time to get an instance of Frequency ?
I already know about __div__, __truediv__, __floordiv__ and other Python special methods, and I already use them in my code to divide instances of classes by numbers or instances of other classes, but I cannot find a way to divide a number by an instance of my class.
Is it possible to implement dividing a number by an instance of a class in Python ?
The __rtruediv__ method is what you're looking for.
When x / y is executed, if type(x) does not implement a __div__(self, other) method where other can be of class type(y), then type(y).__rtruediv__(y, x) is executed, and its result is returned.
Usage:
class Foo:
def __init__(self, x):
self.x = x
def __truediv__(self, other):
return self.x / other
def __rtruediv__(self, other):
return other / self.x
>>> f = Foo(10)
>>> f / 10
1.0
>>> 10 / f
1.0
Yes. You just have to make sure that Time.__rtruediv__() returns a Frequency instance when it receives a float or integer.
Usage:
>>> 100 / Time(2)
Frequency(50.0)
>>> 2.5 / Time(5)
Frequency(0.5)
Implementation:
class Time:
def __init__(self, value):
self.value = value
def __rtruediv__(self, other):
if not isinstance(other, (int, float)):
return NotImplemented
return Frequency(other / self.value)
class Frequency:
def __init__(self, value):
self.value = value
def __repr__(self):
return '{}({})'.format(self.__class__.__name__, self.value)
The python docs contains a full example on implementing the arithmetic operations for your custom classes.
The proper way to handle incompatible types is to return the special value NotImplemented.
NotImplemented
Special value which should be returned by the binary
special methods (e.g. __eq__(), __lt__(), __add__(), __rsub__(), etc.)
to indicate that the operation is not implemented with respect to the
other type
Suppose you try to use a unsupported complex number, returning NotImplemented will eventually cause a TypeError with a correct error message. (at least in python 3)
>>> 100j / Time(2)
Traceback (most recent call last):
File "python", line 1, in <module>
TypeError: unsupported operand type(s) for /: 'complex' and 'Time'
you need to implement __rtruediv__ and__rfloordiv__.
from the documentation
object.__radd__(self, other)
object.__rsub__(self, other)
object.__rmul__(self, other)
object.__rmatmul__(self, other)
object.__rtruediv__(self, other)
object.__rfloordiv__(self, other)
object.__rmod__(self, other)
object.__rdivmod__(self, other)
object.__rpow__(self, other)
object.__rlshift__(self, other)
object.__rrshift__(self, other)
object.__rand__(self, other)
object.__rxor__(self, other)
object.__ror__(self, other)
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 [3] and the
operands are of different types. [4] 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.
Related
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'm currently learning python operator overloading (__radd__ and __add__ to be exact) and I have the following code
class Commuter1:
def __init__(self, val):
self.val = val
def __add__(self, other):
print('add', self.val, other)
return self.val + other
def __radd__(self, other):
print('radd', self.val, other)
return other + self.val
x = Commuter1(88)
y = Commuter1(99)
print(x + y)
I have got the following result
When used separately, I understand how __radd__ and __add__ works. But for the line x + y, I'm not sure why both __radd__ and __add__ methods are evoked.
First, Python looks at the types of x and y to decide whether to call x.__add__ or y.__radd__. Since they're both the same type Commuter1, it tries x.__add__ first.
Then, inside your __add__ method, you do this:
return self.val + other
So, Python looks at the types of self.val and other to decide whether to call self.val.__add__ or other.__radd__. Since they're unrelated types int and Commuter1, it tries int.__add__ first.
But int.__add__ returns NotImplemented for a type it doesn't know about, so Python falls back to calling other.__radd__.
Inside your __radd__ method, you do this:
return other + self.val
So, Python looks at the types of other and self.val to decide whether to call other.__add__ or self.val.__radd__. Since they both the same type int, it tries __add__ first.
And of course int.__add__ works on another int, so it returns a value for the inner + inside your __radd__, which you return, which returns a value for the + inside __add__, which you return, which returns a value for the top-level +, which you print.
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)
For example, if I wanted to apply mathematical operations on objects in the following way:
class A(object):
def __init__(self, value):
self.value = value
def __repr__(self):
return value
assert(A(1) + A(2) == 3)
I am getting the following error: TypeError: unsupported operand type(s) for +: 'A' and 'A'
Is it possible to evaluate objects to primitives so that I can apply simple operations on them? Similarly how you could use implicit conversions in Scala.
You can implement __add__ to define addition on your class.
class A(object):
def __init__(self, value):
self.value = value
def __repr__(self):
return 'A(%r)'%self.value
def __add__(self, other):
return A(self.value+other.value)
>>> A(1)+A(2)
A(3)
This implementation assumes that you are only trying to add instances of A to other instances of A to get a third instance of A. You can write an __add__ adaptable to what type of operand you need it to work for.
See also __radd__ and __iadd__.
That depends on what you're trying to do. You can define the + operator by defining the __add__ method:
class A(object):
def __init__(self, value):
self.value = value
def __repr__(self):
return value
def __add__(self, other):
return A(self.value + other.value)
then of course in your example code you're trying to compare it to an integer which also need to be defined - which is done by implementing the __eq__ method:
def __eq__(self, other):
try:
self.value == other.value
except AttributeError: # other wasn't of class A, try to compare directly instead
return self.value == other
(implicit typecasts on the other hand is not available as far as I know)
The problem is that there isn't enough context in the expression to decide what the objects should be converted to. Python has various methods that can be defined on an object that implement various operators, including the __add__() and __radd__() methods.
There isn't enough context to know that foo should be equivalent to foo.value, so with Python's philosophy explicit is better than implicit. You can certainly subclass int, but then the operators won't produce your new class, and the object itself would remain immutable (as numbers in Python generally are). Notably, ctypes such as c_int32 have a value attribute like your example but do not implement numeric operators.
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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