问题:如何获取Python当前模块中所有类的列表?

我见过很多人从一个模块中提取所有类的示例,通常是这样的:

# foo.py
class Foo:
    pass

# test.py
import inspect
import foo

for name, obj in inspect.getmembers(foo):
    if inspect.isclass(obj):
        print obj

太棒了

但是我无法找到如何从当前模块中获取所有类。

# foo.py
import inspect

class Foo:
    pass

def print_classes():
    for name, obj in inspect.getmembers(???): # what do I do here?
        if inspect.isclass(obj):
            print obj

# test.py
import foo

foo.print_classes()

这可能确实很明显,但是我什么也找不到。谁能帮我吗?

I’ve seen plenty of examples of people extracting all of the classes from a module, usually something like:

# foo.py
class Foo:
    pass

# test.py
import inspect
import foo

for name, obj in inspect.getmembers(foo):
    if inspect.isclass(obj):
        print obj

Awesome.

But I can’t find out how to get all of the classes from the current module.

# foo.py
import inspect

class Foo:
    pass

def print_classes():
    for name, obj in inspect.getmembers(???): # what do I do here?
        if inspect.isclass(obj):
            print obj

# test.py
import foo

foo.print_classes()

This is probably something really obvious, but I haven’t been able to find anything. Can anyone help me out?


回答 0

尝试这个:

import sys
current_module = sys.modules[__name__]

在您的情况下:

import sys, inspect
def print_classes():
    for name, obj in inspect.getmembers(sys.modules[__name__]):
        if inspect.isclass(obj):
            print(obj)

甚至更好:

clsmembers = inspect.getmembers(sys.modules[__name__], inspect.isclass)

因为inspect.getmembers()带谓语。

Try this:

import sys
current_module = sys.modules[__name__]

In your context:

import sys, inspect
def print_classes():
    for name, obj in inspect.getmembers(sys.modules[__name__]):
        if inspect.isclass(obj):
            print(obj)

And even better:

clsmembers = inspect.getmembers(sys.modules[__name__], inspect.isclass)

Because inspect.getmembers() takes a predicate.


回答 1

关于什么

g = globals().copy()
for name, obj in g.iteritems():

What about

g = globals().copy()
for name, obj in g.iteritems():

?


回答 2

我不知道是否有“适当的”方法来执行此操作,但是您的代码片段import foo处在正确的轨道上:只需将其添加到foo.py中,do inspect.getmembers(foo),它就可以正常工作。

I don’t know if there’s a ‘proper’ way to do it, but your snippet is on the right track: just add import foo to foo.py, do inspect.getmembers(foo), and it should work fine.


回答 3

我能够从dir内置plus中获得所需的一切getattr

# Works on pretty much everything, but be mindful that 
# you get lists of strings back

print dir(myproject)
print dir(myproject.mymodule)
print dir(myproject.mymodule.myfile)
print dir(myproject.mymodule.myfile.myclass)

# But, the string names can be resolved with getattr, (as seen below)

虽然,它的确看起来像个毛线球:

def list_supported_platforms():
    """
        List supported platforms (to match sys.platform)

        @Retirms:
            list str: platform names
    """
    return list(itertools.chain(
        *list(
            # Get the class's constant
            getattr(
                # Get the module's first class, which we wrote
                getattr(
                    # Get the module
                    getattr(platforms, item),
                    dir(
                        getattr(platforms, item)
                    )[0]
                ),
                'SYS_PLATFORMS'
            )
            # For each include in platforms/__init__.py 
            for item in dir(platforms)
            # Ignore magic, ourselves (index.py) and a base class.
            if not item.startswith('__') and item not in ['index', 'base']
        )
    ))

I was able to get all I needed from the dir built in plus getattr.

# Works on pretty much everything, but be mindful that 
# you get lists of strings back

print dir(myproject)
print dir(myproject.mymodule)
print dir(myproject.mymodule.myfile)
print dir(myproject.mymodule.myfile.myclass)

# But, the string names can be resolved with getattr, (as seen below)

Though, it does come out looking like a hairball:

def list_supported_platforms():
    """
        List supported platforms (to match sys.platform)

        @Retirms:
            list str: platform names
    """
    return list(itertools.chain(
        *list(
            # Get the class's constant
            getattr(
                # Get the module's first class, which we wrote
                getattr(
                    # Get the module
                    getattr(platforms, item),
                    dir(
                        getattr(platforms, item)
                    )[0]
                ),
                'SYS_PLATFORMS'
            )
            # For each include in platforms/__init__.py 
            for item in dir(platforms)
            # Ignore magic, ourselves (index.py) and a base class.
            if not item.startswith('__') and item not in ['index', 'base']
        )
    ))

回答 4

import pyclbr
print(pyclbr.readmodule(__name__).keys())

请注意,stdlib的Python类浏览器模块使用静态源分析,因此它仅适用于由实际.py文件支持的模块。

import pyclbr
print(pyclbr.readmodule(__name__).keys())

Note that the stdlib’s Python class browser module uses static source analysis, so it only works for modules that are backed by a real .py file.


回答 5

如果要拥有属于当前模块的所有类,则可以使用以下方法:

import sys, inspect
def print_classes():
    is_class_member = lambda member: inspect.isclass(member) and member.__module__ == __name__
    clsmembers = inspect.getmembers(sys.modules[__name__], is_class_member)

如果您使用Nadia的答案并且要在模块上导入其他类,则这些类也将被导入。

因此,这就是为什么member.__module__ == __name__要添加到上使用的谓词的原因is_class_member。该语句检查该类是否确实属于该模块。

谓词是一个函数(可调用),它返回布尔值。

If you want to have all the classes, that belong to the current module, you could use this :

import sys, inspect
def print_classes():
    is_class_member = lambda member: inspect.isclass(member) and member.__module__ == __name__
    clsmembers = inspect.getmembers(sys.modules[__name__], is_class_member)

If you use Nadia’s answer and you were importing other classes on your module, that classes will be being imported too.

So that’s why member.__module__ == __name__ is being added to the predicate used on is_class_member. This statement checks that the class really belongs to the module.

A predicate is a function (callable), that returns a boolean value.


回答 6

另一个可在Python 2和3中使用的解决方案:

#foo.py
import sys

class Foo(object):
    pass

def print_classes():
    current_module = sys.modules[__name__]
    for key in dir(current_module):
        if isinstance( getattr(current_module, key), type ):
            print(key)

# test.py
import foo
foo.print_classes()

Another solution which works in Python 2 and 3:

#foo.py
import sys

class Foo(object):
    pass

def print_classes():
    current_module = sys.modules[__name__]
    for key in dir(current_module):
        if isinstance( getattr(current_module, key), type ):
            print(key)

# test.py
import foo
foo.print_classes()

回答 7

这是我用来获取当前模块中已定义(即未导入)的所有类的行。根据PEP-8,它有点长,但是您可以根据需要进行更改。

import sys
import inspect

classes = [name for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass) 
          if obj.__module__ is __name__]

这为您提供了一个类名列表。如果您想要类对象本身,只需保留obj即可。

classes = [obj for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass)
          if obj.__module__ is __name__]

根据我的经验,这是更有用的。

This is the line that I use to get all of the classes that have been defined in the current module (ie not imported). It’s a little long according to PEP-8 but you can change it as you see fit.

import sys
import inspect

classes = [name for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass) 
          if obj.__module__ is __name__]

This gives you a list of the class names. If you want the class objects themselves just keep obj instead.

classes = [obj for name, obj in inspect.getmembers(sys.modules[__name__], inspect.isclass)
          if obj.__module__ is __name__]

This is has been more useful in my experience.


回答 8

import Foo 
dir(Foo)

import collections
dir(collections)
import Foo 
dir(Foo)

import collections
dir(collections)

回答 9

我认为您可以做这样的事情。

class custom(object):
    __custom__ = True
class Alpha(custom):
    something = 3
def GetClasses():
    return [x for x in globals() if hasattr(globals()[str(x)], '__custom__')]
print(GetClasses())`

如果您需要自己的类

I think that you can do something like this.

class custom(object):
    __custom__ = True
class Alpha(custom):
    something = 3
def GetClasses():
    return [x for x in globals() if hasattr(globals()[str(x)], '__custom__')]
print(GetClasses())`

if you need own classes


回答 10

我经常发现自己在编写命令行实用程序,其中第一个参数旨在引用许多不同类中的一个。例如./something.py feature command —-arguments,where Feature是一个类,并且command是该类上的一个方法。这是一个使这变得容易的基类。

假设该基类与所有子类都位于一个目录中。然后ArgBaseClass(foo = bar).load_subclasses(),您可以拨打电话,这将返回字典。例如,如果目录如下所示:

  • arg_base_class.py
  • feature.py

假设feature.py工具class Feature(ArgBaseClass),则上述调用load_subclasses将返回{ 'feature' : <Feature object> }。相同的kwargsfoo = bar)将传递给Feature该类。

#!/usr/bin/env python3
import os, pkgutil, importlib, inspect

class ArgBaseClass():
    # Assign all keyword arguments as properties on self, and keep the kwargs for later.
    def __init__(self, **kwargs):
        self._kwargs = kwargs
        for (k, v) in kwargs.items():
            setattr(self, k, v)
        ms = inspect.getmembers(self, predicate=inspect.ismethod)
        self.methods = dict([(n, m) for (n, m) in ms if not n.startswith('_')])

    # Add the names of the methods to a parser object.
    def _parse_arguments(self, parser):
        parser.add_argument('method', choices=list(self.methods))
        return parser

    # Instantiate one of each of the subclasses of this class.
    def load_subclasses(self):
        module_dir = os.path.dirname(__file__)
        module_name = os.path.basename(os.path.normpath(module_dir))
        parent_class = self.__class__
        modules = {}
        # Load all the modules it the package:
        for (module_loader, name, ispkg) in pkgutil.iter_modules([module_dir]):
            modules[name] = importlib.import_module('.' + name, module_name)

        # Instantiate one of each class, passing the keyword arguments.
        ret = {}
        for cls in parent_class.__subclasses__():
            path = cls.__module__.split('.')
            ret[path[-1]] = cls(**self._kwargs)
        return ret

I frequently find myself writing command line utilities wherein the first argument is meant to refer to one of many different classes. For example ./something.py feature command —-arguments, where Feature is a class and command is a method on that class. Here’s a base class that makes this easy.

The assumption is that this base class resides in a directory alongside all of its subclasses. You can then call ArgBaseClass(foo = bar).load_subclasses() which will return a dictionary. For example, if the directory looks like this:

  • arg_base_class.py
  • feature.py

Assuming feature.py implements class Feature(ArgBaseClass), then the above invocation of load_subclasses will return { 'feature' : <Feature object> }. The same kwargs (foo = bar) will be passed into the Feature class.

#!/usr/bin/env python3
import os, pkgutil, importlib, inspect

class ArgBaseClass():
    # Assign all keyword arguments as properties on self, and keep the kwargs for later.
    def __init__(self, **kwargs):
        self._kwargs = kwargs
        for (k, v) in kwargs.items():
            setattr(self, k, v)
        ms = inspect.getmembers(self, predicate=inspect.ismethod)
        self.methods = dict([(n, m) for (n, m) in ms if not n.startswith('_')])

    # Add the names of the methods to a parser object.
    def _parse_arguments(self, parser):
        parser.add_argument('method', choices=list(self.methods))
        return parser

    # Instantiate one of each of the subclasses of this class.
    def load_subclasses(self):
        module_dir = os.path.dirname(__file__)
        module_name = os.path.basename(os.path.normpath(module_dir))
        parent_class = self.__class__
        modules = {}
        # Load all the modules it the package:
        for (module_loader, name, ispkg) in pkgutil.iter_modules([module_dir]):
            modules[name] = importlib.import_module('.' + name, module_name)

        # Instantiate one of each class, passing the keyword arguments.
        ret = {}
        for cls in parent_class.__subclasses__():
            path = cls.__module__.split('.')
            ret[path[-1]] = cls(**self._kwargs)
        return ret

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