Default constructor parameters in pyyaml

I encountered the same problem: yaml_tag doesn't work for some reason. So I used alternative approach:

import yaml

def constructor(loader, node) :
    fields = loader.construct_mapping(node)
    return Test(**fields)

yaml.add_constructor('!Test', constructor)

class Test(object) :
    def __init__(self, foo, bar=3) :
        self.foo = foo
        self.bar = bar
    def __repr__(self):
        return "%s(foo=%r, bar=%r)" % (self.__class__.__name__, self.foo, self.bar)

print yaml.load("""
- !Test { foo: 1 }
- !Test { foo: 10, bar: 20 }""")

Output:

[Test(foo=1, bar=3), Test(foo=10, bar=20)]

Based on alexanderlukanin13's answer. Here's my cut.

import yaml

YAMLObjectTypeRegistry = {}

def register_type(target):
    if target.__name__ in YAMLObjectTypeRegistry:
        print "{0} already in registry.".format(target.__name__)
    elif 'yaml_tag' not in target.__dict__.keys():
        print target.__dict__
        raise TypeError("{0} must have yaml_tag attribute".format(
            target.__name__))
    elif target.__dict__['yaml_tag'] is None:
        pass
    else:
        YAMLObjectTypeRegistry[target.__name__] = target
        yaml.add_constructor(
                target.__dict__['yaml_tag'],
                lambda loader, node: target(**loader.construct_mapping(node)))
        print "{0} added to registry.".format(target.__name__)

class RegisteredYAMLObjectType(type):
    def __new__(meta, name, bases, class_dict):
        cls = type.__new__(meta, name, bases, class_dict)
        register_type(cls)
        return cls

class RegisteredYAMLObject(object):
    __metaclass__=RegisteredYAMLObjectType
    yaml_tag = None

You can then use it like this:

class MyType(registry.RegisteredYAMLObject):
    yaml_tag = u'!mytype'
    def __init__(self, name, attr1='default1', attr2='default2'):
        super(MyType, self).__init__()
        self.name = name
        self.attr1 = attr1
        self.attr2 = attr2