为什么numpy.dtype('float64')很特殊?

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) == None):
    print "Surprise!!!!"

谢谢:)

推荐答案

如果要比较python中None的任意对象，则需要使用:

if you want to compare an arbitrary object against exactly None in python you need to use:

object is None

在这种情况下，任何对象都可能会覆盖其比较运算符，从而无法执行您期望的操作.

Like in this case any object may override its comparison operator to not do what you are expecting.

为什么，在dtypes上下文中，dtype('float64')等同于None，就像dtypes等同于typestrings一样

As for why, dtype('float64') is equivalent to None in the context of dtypes in the same way dtypes are equivalent to typestrings

np.dtype('i4') == 'i4'
True

平等不是身份.

关于为什么dtype(None) == dtype('float64')的原因，numpy中的许多函数都具有dtype=None关键字参数.在大多数情况下，这意味着默认dtype为dtype(None).一个示例是np.zeros.但也有例外可以从参数推断出dtype时，例如np.arange(10)的情况，其中默认dtype将是整数类型(我认为是np.intp).

As for why dtype(None) == dtype('float64'), many functions in numpy have dtype=None keyword arguments. In most cases this means default dtype which is dtype(None). An example is np.zeros. But there are exceptions, e.g. when the dtype can be inferred from the arguments, like in the case of np.arange(10) where the default dtype will be of integer type (np.intp I think).

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