scipy.stats.rv_discrete 子类的实例在 pmf() 方法上引发错误 [英] Instance of scipy.stats.rv_discrete subclass throws error on pmf() method

问题描述

我想创建一个 scipy.stats.rv_discrete 的子类来添加一些额外的方法.但是，当我尝试访问子类的 pmf() 方法时，会引发错误.请看下面的例子:

I want to create a subsclass of scipy.stats.rv_discrete to add some additional methods. However, when I try to access the pmf() method of the subclass, an error is raised. Please see the following example:

import numpy as np
from scipy import stats

class sub_rv_discrete(stats.rv_discrete):
  pass

xk = np.arange(2)
pk = (0.5, 0.5)

instance_subclass = sub_rv_discrete(values=(xk, pk))
instance_subclass.pmf(xk)

结果:

Traceback (most recent call last):

  File "<ipython-input-48-129655c38e6a>", line 11, in <module>
    instance.pmf(xk)

  File "C:\Anaconda3\lib\site-packages\scipy\stats\_distn_infrastructure.py", line 2832, in pmf
    args, loc, _ = self._parse_args(*args, **kwds)

AttributeError: 'rv_sample' object has no attribute '_parse_args'

尽管如此，如果我直接使用 stats.rv_discrete，一切都很好:

Despite that, if I use stats.rv_discrete directly, everything is fine:

instance_class = stats.rv_discrete(values=(xk, pk))
instance_class.pmf(xk)

---> array([ 0.5,  0.5])

推荐答案

@josef-pkt 对 github 如下:

通过常规的子类化创建一个 rv_sample 类并且不会初始化正确的类

going through the regular subclassing creates a rv_sample class and doesn't init the correct class

以下适用于 0.18.1(我现在已经打开)

the following works for me with 0.18.1 (which I have open right now)

from scipy.stats._distn_infrastructure import rv_sample

class subc_rv_discrete(rv_sample):

    def __new__(cls, *args, **kwds):
        return super(subc_rv_discrete, cls).__new__(cls)

xk = [0,1,2,3]
pk = [0.1, 0.2, 0.3, 0.4]

inst = subc_rv_discrete(values=(xk, pk))
print(inst.pmf(xk))
print(inst.__class__)

也许这将在进一步的 scipy 版本中修复...

maybe this will be fixed in further scipy releases...

这篇关于scipy.stats.rv_discrete 子类的实例在 pmf() 方法上引发错误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IT屋！