Python: max/min builtin functions depend on parameter order
Solution 1:
In [19]: 1>float('nan')
Out[19]: False
In [20]: float('nan')>1
Out[20]: False
The float nan is neither bigger nor smaller than the integer 1.
max starts by choosing the first element, and only replaces it when it finds an element which is strictly larger.
In [31]: max(1,float('nan'))
Out[31]: 1
Since nan is not larger than 1, 1 is returned.
In [32]: max(float('nan'),1)
Out[32]: nan
Since 1 is not larger than nan, nan is returned.
PS. Note that np.max treats float('nan') differently:
In [36]: import numpy as np
In [91]: np.max([1,float('nan')])
Out[91]: nan
In [92]: np.max([float('nan'),1])
Out[92]: nan
but if you wish to ignore np.nans, you can use np.nanmax:
In [93]: np.nanmax([1,float('nan')])
Out[93]: 1.0
In [94]: np.nanmax([float('nan'),1])
Out[94]: 1.0
Solution 2:
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.
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
Solution 3:
Max works the following way:
The first item is set as maxval and then the next is compared to this value. The comparation will always return False:
>>> float('nan') < 1
False
>>> float('nan') > 1
False
So if the first value is nan, then (since the comparation returns false) it will not be replaced upon the next step.
OTOH if 1 is the first, the same happens: but in this case, since 1 was set, it will be the maximum.
You can verify this in the python code, just look up the function min_max in Python/bltinmodule.c