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257

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3

This would be similar to the java.lang.Object.hashcode() method.

I need to store objects I have no control over in a set, and make sure that only if two objects are actually the same object (not contain the same values) will the values be overwritten.

+3  A: 
id(x)

will do the trick for you. But I'm curious, what's wrong about the set of objects (which does combine objects by value)?

For your particular problem I would probably keep the set of ids or of wrapper objects. A wrapper object will contain one reference and compare by x==y <==> x.ref is y.ref.

It's also worth noting that Python objects have a hash function as well. This function is necessary to put an object into a set or dictionary. It is supposed to sometimes collide for different objects, though good implementations of hash try to make it less likely.

ilya n.
+2  A: 

That's what "is" is for.

Instead of testing "if a == b", which tests for the same value,

test "if a is b", which will test for the same identifier.

Vicki Laidler
Would it be accurate to say that Python's "is" is like Java's "==" and Python's "==" is like Java's "equals()"?
MatrixFrog
@MatrixFrog Yes.
Imagist
+1  A: 

As ilya n mentions, id(x) produces a unique identifier for an object.

But your question is confusing, since Java's hashCode method doesn't give a unique identifier. Java's hashCode works like most hash functions: it always returns the same value for the same object, two objects that are equal always get equal codes, and unequal hash values imply unequal hash codes. In particular, two different and unequal objects can get the same value.

This is confusing because cryptographic hash functions are quite different from this, and more like (though not exactly) the "unique id" that you asked for.

The Python equivalent of Java's hashCode method is hash(x).

Ned Batchelder
That's why I explicitly mentioned java.lang.Object's hashcode method. Which, by default, produces a unique long for that object. I realize the other implementations of hashcode can, and do, differ.
oneself
But that's the point: hashCode doesn't produce a unique long. It produces longs which try hard not to collide, but you can't map strings to longs and produce a unique value.
Ned Batchelder