为什么numpy.dtype(numpy.float64)评估为False [英] Why does numpy.dtype(numpy.float64) evaluate to False

问题描述

有人可以解释以下脚本输出背后的逻辑吗?

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

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

特别考虑:

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

谢谢:)

推荐答案

np.dtype没有定义__nonzero__，但确实定义了__len__.根据文档，这意味着当您在布尔上下文中使用它时，如果__len__返回非零值，则将得出True.但无论您传入哪种类型，它始终返回零:

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

另一方面，复合数据类型的确返回非零，因此为True:

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

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

然后您可能会问，当具有单个元素的复合物1的长度为1时，为什么简单dtype的长度"为零.我想这与维数有关:具有简单dtype和一维大小的数组本身就是一维的，但是具有复合dtype和一维大小的数组可能被认为是二维的，而不管有多少个元素属于复合dtype.

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.

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