是否有一种方法可以获取类实例上存在的属性列表?

class new_class():
    def __init__(self, number):
        self.multi = int(number) * 2
        self.str = str(number)

a = new_class(2)
print(', '.join(a.SOMETHING))

期望的结果是输出"multi, str"。我希望它能看到脚本各个部分的当前属性。


当前回答

除了这些答案之外,我还将包括一个函数(python 3),用于输出任何值的几乎整个结构。它使用dir来建立属性名的完整列表,然后对每个名称使用getattr。它会显示值的每个成员的类型,如果可能的话还会显示整个成员:

import json

def get_info(obj):

  type_name = type(obj).__name__
  print('Value is of type {}!'.format(type_name))
  prop_names = dir(obj)

  for prop_name in prop_names:
    prop_val = getattr(obj, prop_name)
    prop_val_type_name = type(prop_val).__name__
    print('{} has property "{}" of type "{}"'.format(type_name, prop_name, prop_val_type_name))

    try:
      val_as_str = json.dumps([ prop_val ], indent=2)[1:-1]
      print('  Here\'s the {} value: {}'.format(prop_name, val_as_str))
    except:
      pass

现在,以下任何一项都应该让你有所了解:

get_info(None)
get_info('hello')

import numpy
get_info(numpy)
# ... etc.

其他回答

获取对象的属性

class new_class():
    def __init__(self, number):
    self.multi = int(number) * 2
    self.str = str(number)

new_object = new_class(2)                
print(dir(new_object))                   #total list attributes of new_object
attr_value = new_object.__dict__         
print(attr_value)                        #Dictionary of attribute and value for new_class                   

for attr in attr_value:                  #attributes on  new_class
    print(attr)

输出

['__class__', '__delattr__', '__dict__', '__dir__', '__doc__','__eq__', '__format__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__', '__init_subclass__', '__le__', '__lt__', '__module__', '__ne__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__', 'multi', 'str']

{'multi': 4, 'str': '2'}

multi
str

你要这个干什么?如果不知道你的确切意图,可能很难得到最好的答案。

It is almost always better to do this manually if you want to display an instance of your class in a specific way. This will include exactly what you want and not include what you don't want, and the order will be predictable. If you are looking for a way to display the content of a class, manually format the attributes you care about and provide this as the __str__ or __repr__ method for your class. If you want to learn about what methods and such exist for an object to understand how it works, use help. help(a) will show you a formatted output about the object's class based on its docstrings. dir exists for programatically getting all the attributes of an object. (Accessing __dict__ does something I would group as the same but that I wouldn't use myself.) However, this may not include things you want and it may include things you do not want. It is unreliable and people think they want it a lot more often than they do. On a somewhat orthogonal note, there is very little support for Python 3 at the current time. If you are interested in writing real software you are going to want third-party stuff like numpy, lxml, Twisted, PIL, or any number of web frameworks that do not yet support Python 3 and do not have plans to any time too soon. The differences between 2.6 and the 3.x branch are small, but the difference in library support is huge.

除了这些答案之外,我还将包括一个函数(python 3),用于输出任何值的几乎整个结构。它使用dir来建立属性名的完整列表,然后对每个名称使用getattr。它会显示值的每个成员的类型,如果可能的话还会显示整个成员:

import json

def get_info(obj):

  type_name = type(obj).__name__
  print('Value is of type {}!'.format(type_name))
  prop_names = dir(obj)

  for prop_name in prop_names:
    prop_val = getattr(obj, prop_name)
    prop_val_type_name = type(prop_val).__name__
    print('{} has property "{}" of type "{}"'.format(type_name, prop_name, prop_val_type_name))

    try:
      val_as_str = json.dumps([ prop_val ], indent=2)[1:-1]
      print('  Here\'s the {} value: {}'.format(prop_name, val_as_str))
    except:
      pass

现在,以下任何一项都应该让你有所了解:

get_info(None)
get_info('hello')

import numpy
get_info(numpy)
# ... etc.

上面所有的答案都是正确的,你有三个选项可以选择

dir () vars () __dict__

>>> dir(a)
['__class__', '__delattr__', '__dict__', '__dir__', '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__', '__init_subclass__', '__le__', '__lt__', '__module__', '__ne__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__', 'multi', 'str']
>>> vars(a)
{'multi': 4, 'str': '2'}
>>> a.__dict__
{'multi': 4, 'str': '2'}

人们经常提到,要列出一个完整的属性列表,应该使用dir()。但是请注意,与普遍观点相反,dir()并没有显示所有属性。例如,你可能会注意到__name__可能在类的dir()列表中缺失,即使你可以从类本身访问它。从dir()的文档(Python 2, Python 3):

因为提供dir()主要是为了方便在 交互式提示符,它试图提供一组有趣的名称 它不仅仅是试图提供一个严格或一致定义的集合 的名称,其详细行为可能在不同版本之间更改。为 属性时,元类属性不在结果列表中 参数是一个类。

像下面这样的函数往往更完整,尽管不能保证完整性,因为dir()返回的列表可能受到许多因素的影响,包括实现__dir__()方法,或在类或其父类之一上自定义__getattr__()或__getattribute__()。详情请参阅所提供的链接。

def dirmore(instance):
    visible = dir(instance)
    visible += [a for a in set(dir(type)).difference(visible)
                if hasattr(instance, a)]
    return sorted(visible)