I want to write a custom class that behaves like dict
- so, I am inheriting from dict
.
My question, though, is: Do I need to create a priva
I really don't see the right answer to this anywhere
class MyClass(dict):
def __init__(self, a_property):
self[a_property] = a_property
All you are really having to do is define your own __init__
- that really is all that there is too it.
Another example (little more complex):
class MyClass(dict):
def __init__(self, planet):
self[planet] = planet
info = self.do_something_that_returns_a_dict()
if info:
for k, v in info.items():
self[k] = v
def do_something_that_returns_a_dict(self):
return {"mercury": "venus", "mars": "jupiter"}
This last example is handy when you want to embed some kind of logic.
Anyway... in short class GiveYourClassAName(dict)
is enough to make your class act like a dict. Any dict operation you do on self
will be just like a regular dict.
Don’t inherit from Python built-in dict, ever! for example update
method woldn't use __setitem__
, they do a lot for optimization. Use UserDict.
from collections import UserDict
class MyDict(UserDict):
def __delitem__(self, key):
pass
def __setitem__(self, key, value):
pass
class Mapping(dict):
def __setitem__(self, key, item):
self.__dict__[key] = item
def __getitem__(self, key):
return self.__dict__[key]
def __repr__(self):
return repr(self.__dict__)
def __len__(self):
return len(self.__dict__)
def __delitem__(self, key):
del self.__dict__[key]
def clear(self):
return self.__dict__.clear()
def copy(self):
return self.__dict__.copy()
def has_key(self, k):
return k in self.__dict__
def update(self, *args, **kwargs):
return self.__dict__.update(*args, **kwargs)
def keys(self):
return self.__dict__.keys()
def values(self):
return self.__dict__.values()
def items(self):
return self.__dict__.items()
def pop(self, *args):
return self.__dict__.pop(*args)
def __cmp__(self, dict_):
return self.__cmp__(self.__dict__, dict_)
def __contains__(self, item):
return item in self.__dict__
def __iter__(self):
return iter(self.__dict__)
def __unicode__(self):
return unicode(repr(self.__dict__))
o = Mapping()
o.foo = "bar"
o['lumberjack'] = 'foo'
o.update({'a': 'b'}, c=44)
print 'lumberjack' in o
print o
In [187]: run mapping.py
True
{'a': 'b', 'lumberjack': 'foo', 'foo': 'bar', 'c': 44}