[Python] Numpy min of a function returns the function it self

Okey I don't even know where to ask this or what I am exactly asking, but is basically what the title says: if the argument of numpy.min is a function, it doesn't matter if its defined by the user or not, the output is such function.

Let me put an example:

import numpy as np

def get_third_value(x):
return x[2]

xx = np.arange(10)

print(np.min(get_third_value)) # output: <function the_third at 0x7ffb58c2b240>
print(np.min(max)) # output: <built-in function max>

# you can even use them:
print(np.min(max)(xx)) # output : 9
print(np.min(get_third_value)(xx)) # output: 2


Why this happens? Is this the intended behavior? If so, why?

I have checked other numpy functions and the same happens in some of them

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