Overload python ternary operator

Question

Is it possible to overload the ternary operator in python? Basically what I want is something like:

class A(object):
    def __ternary__(self, a, c):
        return a + c

a = A()
print "asdf" if a else "fdsa" # prints "asdffdsa"

I'm trying to implement a symbolic package and basically want something that can do things like:

sym = Symbol("s")
result = 1 if sym < 3 else 10
print result.evaluate(sym=2) # prints 1
print result.evaluate(sym=4) # prints 10

Edit: Let me put out a bit more complex example to show how this could be layered upon.

sym = Symbol("s")
result = 1 if sym < 3 else 10
...
something_else = (result+1)*3.5
...
my_other_thing = sqrt(something_else)

print my_other_thing.evaluate(sym=2) # prints sqrt(7) or rather the decimal equivalent

The point is, I don't need to just be able to late evaluate the one ternary operator, I need to take the result and do other symbolic stuff with that before finally evaluating. Furthermore, my code can do partial evaluations where I give it a few bindings and it returns another symbolic expression if it can't evaluate the full expression.

My backup plan is just to directly use the ternary class taking 3 expressions objects that I would need to make anyway. I was just trying to hide the generation of this class with an operator overload. Basically:

a = TernaryOperator(a,b,c)
# vs
b = a if b else c

Wouldn't it be better to overload the comparison operators (and convert-to-bool operator; don't know what the proper Python term is) instead? — Aasmund Eldhuset
– Aasmund Eldhuset, Commented Feb 18, 2015 at 3:24
I think you have the comparison backwards on your example - should be if sym < 3 — Hugh Bothwell
– Hugh Bothwell, Commented Feb 18, 2015 at 3:27
@AasmundEldhuset How would you implement what I'm trying to do then? If I understand correct, what you're suggesting would get evaluated at the result=... line instead of the print lines. — CrazyCasta
– CrazyCasta, Commented Feb 18, 2015 at 3:30

Hugh Bothwell · Accepted Answer · 2015-02-18 03:36:32Z

1

look at the sympy module; it already does this
for simple comparison, write A.__eq__ and A.__lt__ methods and use the total_ordering class decorator; this should be sufficient for comparing two As or an A and a constant

write it as a lambda,

result = lambda sym: 1 if sym < 3 else 10
print(result(2))    # => 1
print(result(4))    # => 10

answered Feb 18, 2015 at 3:36

Hugh Bothwell

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2 Comments

Aasmund Eldhuset Over a year ago

Your lambda operates on the integers 2 and 4.

Hugh Bothwell Over a year ago

Yes, it does; per his example, result.evaluate(sym=2), this may be sufficient for his needs.

Aasmund Eldhuset · Accepted Answer · 2015-02-18 03:36:15Z

0

Overload the comparison operators instead (something you probably needed to do anyway):

class A(object):
    def __lt__(self, other):
        return self.value() < other.value() # Replace with your own implementation of <

Then, use lambda functions to achieve the delayed evaluation you desire:

sym = Symbol("s")
result = lambda s: 1 if s < 3 else 10
sym.set(2)
print result(sym) # prints 1
sym.set(4)
print result(sym) # prints 10

(I don't think you can overload the assignment operator, as it doesn't actually perform an operation on any object, but rather on a variable.)

answered Feb 18, 2015 at 3:36

Aasmund Eldhuset

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1 Comment

CrazyCasta Over a year ago

it's actually the ternary operator I was hoping to overload (not the assignment operator) but I have a feeling that you just can't overload it, which is too bad

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