问题描述
我今天在python if子句中遇到了意外的结果:
I ran into unexpected results in a python if clause today:
import numpy
if numpy.allclose(6.0, 6.1, rtol=0, atol=0.5):
print 'close enough' # works as expected (prints message)
if numpy.allclose(6.0, 6.1, rtol=0, atol=0.5) is True:
print 'close enough' # does NOT work as expected (prints nothing)
经过一番摸索(例如，此问题，尤其是)，我理解了原因:numpy.allclose()返回的type是numpy.bool_而不是普通的bool，并且显然是foo = numpy.bool_(1) ，则if foo的评估结果为True，而if foo is True的评估结果为False.这似乎是is操作员的工作.
After some poking around (i.e., this question, and in particular this answer), I understand the cause: the type returned by numpy.allclose() is numpy.bool_ rather than plain old bool, and apparently if foo = numpy.bool_(1), then if foo will evaluate to True while if foo is True will evaluate to False. This appears to be the work of the is operator.
我的问题是:为什么numpy具有自己的布尔类型，鉴于这种情况，最佳实践是什么?在上面的示例中，我可以写if foo:来获得预期的行为，但是我喜欢更严格的if foo is True:，因为它从返回True中排除了2和[2]之类的东西，有时还不包括显式类型检查是可取的.
My questions are: why does numpy have its own boolean type, and what is best practice in light of this situation? I can get away with writing if foo: to get expected behavior in the example above, but I like the more stringent if foo is True: because it excludes things like 2 and [2] from returning True, and sometimes the explicit type check is desirable.
推荐答案
空间和速度. Numpy将事物存储在紧凑的数组中；如果它可以将布尔值放入单个字节中，则将尝试.使用Python对象很难做到这一点，因为您必须存储引用，这会大大降低计算速度.
Space and speed. Numpy stores things in compact arrays; if it can fit a boolean into a single byte it'll try. You can't easily do this with Python objects, as you have to store references which slows calculations down significantly.
好吧，不要那样做.
这篇关于python vs numpy中的布尔值和类型检查的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！