Home >Backend Development >Python Tutorial >How do you ensure consistent and robust equivalence comparisons for custom Python classes, especially when dealing with subclasses and sets?
In Python, custom classes can implement the __eq__ and __ne__ methods to define equivalence for the == and != operators, respectively. While the traditional method of comparing dictionary attributes is straightforward, it presents certain limitations.
Consider the following scenario:
class Number: def __init__(self, number): self.number = number n1 = Number(1) n2 = Number(1) # Default comparison fails: they are different objects assert n1 != n2
To address this issue, we can override the __eq__ method:
class Number: def __init__(self, number): self.number = number def __eq__(self, other): if isinstance(other, Number): return self.number == other.number return False
However, for Python 2, we also need to implement __ne__ to ensure commutative behavior:
class Number: def __init__(self, number): self.number = number def __eq__(self, other): if isinstance(other, Number): return self.number == other.number return False def __ne__(self, other): return not self.__eq__(other)
This ensures that n1 == n2 evaluates to True, as expected.
Introducing subclasses complicates equivalence comparison:
class SubNumber(Number): pass n3 = SubNumber(1) # Subclass comparison fails for classic-style classes assert n1 == n3 # False (for classic-style classes) assert n3 == n1 # True # Non-commutative comparison assert n1 != n3 # True (for classic-style classes) assert n3 != n1 # False
For classic-style classes, the comparison method is invoked based on the first operand's type, leading to non-commutative behavior. To address this, we can return NotImplemented for unsupported operand types, which delegates the comparison to the other operand's method:
def __eq__(self, other): if isinstance(other, Number): return self.number == other.number return NotImplemented
Lastly, note that sets use object identifiers for hashing, which can lead to incorrect results:
assert len(set([n1, n2, n3])) == 3 # Incorrectly reports 3 unique numbers
To resolve this, we can override the __hash__ method:
def __hash__(self): return hash(tuple(sorted(self.__dict__.items())))
With these enhancements, the equivalence and uniqueness behaviors become correct and consistent, ensuring robust comparisons and accurate representation in sets:
assert len(set([n1, n2, n3])) == 1 assert len(set([n1, n2, n3, n4])) == 2
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