math - 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 the answers.

max raises an exception when the iterable is empty. Why wouldn't Python's max raise an exception when nan is present? Or at least do something useful, like return nan or ignore nan. The current behavior is very unsafe and seems completely unreasonable.

I found an even more surprising consequence of this behavior, so I just posted a related question.

Answer 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.

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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