sup*_*kar 10 python python-attrs
我有类似的东西:
from attr import attrs, attrib
@attrs
class Foo():
max_count = attrib()
@property
def get_max_plus_one(self):
return self.max_count + 1
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现在当我这样做时:
f = Foo(max_count=2)
f.get_max_plus_one =>3
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我想将其转换为字典:
{'max_count':2, 'get_max_plus_one': 3}
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当我使用时,attr.asdict(f)
我没有得到@property
. 我只得到
{'max_count':2}
.
实现上述目标的最干净的方法是什么?
dir
对于这种情况,您可以在对象上使用,并仅获取不以__
ie 开头的属性,忽略魔术方法:
In [496]: class Foo():
...: def __init__(self):
...: self.max_count = 2
...: @property
...: def get_max_plus_one(self):
...: return self.max_count + 1
...:
In [497]: f = Foo()
In [498]: {prop: getattr(f, prop) for prop in dir(f) if not prop.startswith('__')}
Out[498]: {'get_max_plus_one': 3, 'max_count': 2}
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要处理不以 开头的常规方法__
,您可以添加一个callable
测试:
In [521]: class Foo():
...: def __init__(self):
...: self.max_count = 2
...: @property
...: def get_max_plus_one(self):
...: return self.max_count + 1
...: def spam(self):
...: return 10
...:
In [522]: f = Foo()
In [523]: {prop: getattr(f, prop) for prop in dir(f) if not (prop.startswith('__') or callable(getattr(Foo, prop, None)))}
Out[523]: {'get_max_plus_one': 3, 'max_count': 2}
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