Why is numpy.dtype('float64') special?

Question

Can someone explain the logic behind the output of the following script?

import numpy
if(numpy.dtype(numpy.float64) == None):
    print \"Surprise!!!!\"
         


        

Answer 1

Looks like an unfortunate accident: someone decided that dtype(None) would "default" to float (though dtype() is an error). Then someone else wrote dtype.__eq__ such that it converts its second argument to a dtype before comparing. So dtype(float) == None is dtype(float) == dtype(None) which is true.

You can see a comment in the source code here: descriptor.c#L1217

• Get typenum from an object -- None goes to NPY_DEFAULT_TYPE

And of course NPY_DEFAULT_TYPE is float (at least usually).

As for the __eq__ operator, it's here: descriptor.c#L3317. It does what I outlined:

if (!PyArray_DescrCheck(other)) {
    if (PyArray_DescrConverter(other, &new) == NPY_FAIL) {
        return NULL;
    }
}

So that's a conversion from whatever is on the right-hand side of == to a dtype object, via the converter function mentioned before, which turns None into dtype(float).

Edit: I found this pretty interesting and it seems like an accident, so I created a patch and submitted to the maintainers: https://github.com/numpy/numpy/pull/4532 .