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My gut reaction is that both NaN == NaN and NaN != NaN should be false, it to put it another way, NaN != NaN returns True was a surprise to me. Does Numpy do t

• iNic 1y ago
  Yes, in numpy we also have that `np.float64(nan) != np.float64(nan)` evaluates to true.
• nyrikki 1y ago
  In a perfect world, in my opinion is that they are incompatibles, and the equality operation should return False in both cases. But equality is a very complicated concept. I guess if the values are incomparable != preserves PEM But Boolean algebra is a lattice, with two+ binary operators were two of those are meet and join with a shared absorbing property. X == not not X being PEM, we lose that in NP vs co-NP and in vector clocks etc… But that is just my view.
• Pinus 1y ago
  For the built-in float type in Python, the behaviour is a bit funny: >>> nan=float('nan') >>> a=[nan] >>> nan in a True >>> any(nan==x for x in a) False (Because the `in` operator assumes that identity implies equality...)
  • dragonwriter 1y ago
    Well no, the in operator is just defined to produce results equivalent equivalent to any(nan is x or nan==x for x in a); it is counterintuitive to the extent people assume that identity implies equality, but the operator doesn't assume that identity implies equality, it is defined as returning True if either is satisfied. [0] Well, more precisely, this is how the operator behaves for most built in collections; other types can define how it behaves for them by implementing a __contains__() method with the desired semantics. [0] https://docs.python.org/3/reference/expressions.html#membership-test-details https://docs.python.org/3/reference/expressions.html#members...