Python: max/min builtin functions depend on parameter order

Question

max(float(\'nan\'), 1) evaluates to nan

max(1, float(\'nan\')) evaluates to 1

Is it the intended behavior?

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Thanks for

Answer 1

I haven't seen this before, but it makes sense. Notice that nan is a very weird object:

>>> x = float('nan')
>>> x == x
False
>>> x > 1
False
>>> x < 1
False

I would say that the behaviour of max is undefined in this case -- what answer would you expect? The only sensible behaviour is to assume that the operations are antisymmetric.

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Notice that you can reproduce this behaviour by making a broken class:

>>> class Broken(object):
...     __le__ = __ge__ = __eq__ = __lt__ = __gt__ = __ne__ =
...     lambda self, other: False
...
>>> x = Broken()
>>> x == x
False
>>> x < 1
False
>>> x > 1
False
>>> max(x, 1)
<__main__.Broken object at 0x024B5B50>
>>> max(1, x)
1