将模块中的所有类初始化为列表中的无名对象 [英] Initialize all the classes in a module into nameless objects in a list

问题描述

是否可以将python模块中的所有类初始化为无名对象列表?

Is there a way to initialize all classes from a python module into a list of nameless objects?

示例 我有一个模块rules，其中包含类Rule中的所有子类.因此，我确定他们都将实现方法run()并将具有属性name，该属性将在调用__init__

Example I have a module rules which contains all child classes from a class Rule. Because of that, I'm certain they will all implement a method run() and will have a attribute name which will be generated during the call of __init__

我想要一个从这些类动态初始化的对象的列表.通过动态初始化，我的意思是它们不必显式命名.

I would like to have a list of objects dynamically initiated from those classes. By dynamically initialized i mean that they don't have to be named explicitly.

问题是:

是否可以遍历模块中的所有类?

Is it possible to iterate through all classes in a module?

可以启动无名对象吗?

推荐答案

您至少可以采用两种方法.您可以通过调用类的__subclasses__()方法来获取类的所有子类.因此，如果您的父类称为Rule，则可以调用:

There are at least two approaches you can take. You can get all of the subclasses of a class by calling a class's __subclasses__() method. So if your parent class is called Rule, you could call:

rule_list = [cls() for cls in Rule.__subclasses__()]

这将为您提供Rule的所有子类，并且不会将它们限制为在特定模块中找到的子类.

This will give you all subclasses of Rule, and will not limit them to the ones found in a particular module.

如果您有模块的句柄，则可以遍历其内容.像这样:

If you have a handle to your module, you can iterate over its content. Like so:

import rule
rule_list = []
for name in dir(rule):
    value = getattr(rule, name)
    if isinstance(value, type) and issubclass(value, Rule):
        rule_list.append(value())

不幸的是，如果给它一个不是类的对象作为第一个参数，则issubclass会抛出TypeError.因此，您必须以某种方式处理该问题.

Unfortunately, issubclass throws TypeError if you give it an object that is not a class as its first argument. So you have to handle that somehow.

按照@Blckknght的建议处理issubclass怪癖.

dealing with the issubclass quirk per @Blckknght's suggestion.

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