max(float('nan'), 1) evaluates to nan
max(1, float('nan')) evaluates to 1
Is it the intended behavior?
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