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pythonnumpy
Troels Blum
Troels Blum

Reputation: 855

Why does numpy.dtype(numpy.float64) evaluate to False

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

import numpy
if(numpy.dtype(numpy.float64):
    print "Expected"
else:
    print "Surprise!!!!"

Especially considering:

import numpy
if(object):
    print "Expected!"
else:
    print "Surprise"

Thanks :)

Upvotes: 1

Views: 670

Answers (1)

John Zwinck
John Zwinck

Reputation: 249113

np.dtype does not define __nonzero__, but it does define __len__. As per the documentation, this means when you use it in a boolean context, it will evaluate to True if __len__ returns non-zero. But it always returns zero, regardless of what type you pass in:

>>> bool(np.dtype(int))
False
>>> bool(np.dtype(float))
False
>>> bool(np.dtype(np.int8))
False

On the other hand, a compound data type does return nonzero, thus True:

>>> bool(np.dtype([('foo', int)]))
True

You might then ask why the "length" of a simple dtype is zero, when the length of a compound one with a single element is one. I imagine that's something about dimensionality: an array with a simple dtype and one-dimensional size is itself one-dimensional, but an array with a compound dtype and one-dimensional size may be thought of as two-dimensional, regardless of how many elements are in the compound dtype.

Upvotes: 3

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