["python"]
In python, how do I cast a class object to a dict

Let's say I've got a simple class in python class Wharrgarbl(object): def __init__(self, a, b, c, sum, version='old'): self.a = a self.b = b self.c = c self.sum = 6 self.version = version def __int__(self): return self.sum + 9000 def __what_goes_here__(self): return {'a': self.a, 'b': self.b, 'c': self.c} I can cast it to an integer very easily >>> w = Wharrgarbl('one', 'two', 'three', 6) >>> int(w) 9006 Which is great! But, now I want to cast it to a dict in a similar fashion >>> w = Wharrgarbl('one', 'two', 'three', 6) >>> dict(w) {'a': 'one', 'c': 'three', 'b': 'two'} What do I need to define for this to work? I tried substituting both __dict__ and dict for __what_goes_here__, but dict(w) resulted in a TypeError: Wharrgarbl object is not iterable in both cases. I don't think simply making the class iterable will solve the problem. I also attempted many googles with as many different wordings of "python cast object to dict" as I could think of but couldn't find anything relevant :{ Also! Notice how calling w.__dict__ won't do what I want because it's going to contain w.version and w.sum. I want to customize the cast to dict in the same way that I can customize the cast to int by using def int(self). I know that I could just do something like this >>> w.__what_goes_here__() {'a': 'one', 'c': 'three', 'b': 'two'} But I am assuming there is a pythonic way to make dict(w) work since it is the same type of thing as int(w) or str(w). If there isn't a more pythonic way, that's fine too, just figured I'd ask. Oh! I guess since it matters, this is for python 2.7, but super bonus points for a 2.4 old and busted solution as well. There is another question Overloading __dict__() on python class that is similar to this one but may be different enough to warrant this not being a duplicate. I believe that OP is asking how to cast all the data in his class objects as dictionaries. I'm looking for a more customized approach in that I don't want everything in __dict__ included in the dictionary returned by dict(). Something like public vs private variables may suffice to explain what I'm looking for. The objects will be storing some values used in calculations and such that I don't need/want to show up in the resulting dictionaries. UPDATE: I've chosen to go with the asdict route suggested but it was a tough choice selecting what I wanted to be the answer to the question. Both @RickTeachey and @jpmc26 provided the answer I'm going to roll with but the former had more info and options and landed on the same result as well and was upvoted more so I went with it. Upvotes all around though and thanks for the help. I've lurked long and hard on stackoverflow and I'm trying to get my toes in the water more.
There are at least five six ways. The preferred way depends on what your use case is. Option 1: Simply add an asdict() method. Based on the problem description I would very much consider the asdict way of doing things suggested by other answers. This is because it does not appear that your object is really much of a collection: class Wharrgarbl(object): ... def asdict(self): return {'a': self.a, 'b': self.b, 'c': self.c} Using the other options below could be confusing for others unless it is very obvious exactly which object members would and would not be iterated or specified as key-value pairs. Option 1a: Inherit your class from 'typing.NamedTuple' (or the mostly equivalent 'collections.namedtuple'), and use the _asdict method provided for you. from typing import NamedTuple class Wharrgarbl(NamedTuple): a: str b: str c: str sum: int = 6 version: str = 'old' Using a named tuple is a very convenient way to add lots of functionality to your class with a minimum of effort, including an _asdict method. However, a limitation is that, as shown above, the NT will include all the members in its _asdict. If there are members you don't want to include in your dictionary, you'll need to specify which members you want the named tuple _asdict result to include. To do this, you could either inherit from a base namedtuple class using the older collections.namedtuple API: from collections import namedtuple as nt class Wharrgarbl(nt("Basegarble", "a b c")): # note that the typing info below isn't needed for the old API a: str b: str c: str sum: int = 6 version: str = 'old' ...or you could create a base class using the newer API, and inherit from that, using only the dictionary members in the base class: from typing import NamedTuple class Basegarbl(NamedTuple): a: str b: str c: str class Wharrgarbl(Basegarbl): sum: int = 6 version: str = 'old' Another limitation is that NT is read-only. This may or may not be desirable. Option 2: Implement __iter__. Like this, for example: def __iter__(self): yield 'a', self.a yield 'b', self.b yield 'c', self.c Now you can just do: dict(my_object) This works because the dict() constructor accepts an iterable of (key, value) pairs to construct a dictionary. Before doing this, ask yourself the question whether iterating the object as a series of key,value pairs in this manner- while convenient for creating a dict- might actually be surprising behavior in other contexts. E.g., ask yourself the question "what should the behavior of list(my_object) be...?" Additionally, note that accessing values directly using the get item obj["a"] syntax will not work, and keyword argument unpacking won't work. For those, you'd need to implement the mapping protocol. Option 3: Implement the mapping protocol. This allows access-by-key behavior, casting to a dict without using __iter__, and also provides two types of unpacking behavior: mapping unpacking behavior: {**my_obj} keyword unpacking behavior, but only if all the keys are strings: dict(**my_obj) The mapping protocol requires that you provide (at minimum) two methods together: keys() and __getitem__. class MyKwargUnpackable: def keys(self): return list("abc") def __getitem__(self, key): return dict(zip("abc", "one two three".split()))[key] Now you can do things like: >>> m=MyKwargUnpackable() >>> m["a"] 'one' >>> dict(m) # cast to dict directly {'a': 'one', 'b': 'two', 'c': 'three'} >>> dict(**m) # unpack as kwargs {'a': 'one', 'b': 'two', 'c': 'three'} As mentioned above, if you are using a new enough version of python you can also unpack your mapping-protocol object into a dictionary comprehension like so (and in this case it is not required that your keys be strings): >>> {**m} {'a': 'one', 'b': 'two', 'c': 'three'} Note that the mapping protocol takes precedence over the __iter__ method when casting an object to a dict directly (without using kwarg unpacking, i.e. dict(m)). So it is possible- and might be sometimes convenient- to cause the object to have different behavior when used as an iterable (e.g., list(m)) vs. when cast to a dict (dict(m)). But note also that with regular dictionaries, if you cast to a list, it will give the KEYS back, and not the VALUES as you require. If you implement another nonstandard behavior for __iter__ (returning values instead of keys), it could be surprising for other people using your code unless it is very obvious why this would happen. EMPHASIZED: Just because you CAN use the mapping protocol, does NOT mean that you SHOULD do so. Does it actually make sense for your object to be passed around as a set of key-value pairs, or as keyword arguments and values? Does accessing it by key- just like a dictionary- really make sense? Would you also expect your object to have other standard mapping methods such as items, values, get? Do you want to support the in keyword and equality checks (==)? If the answer to these questions is yes, it's probably a good idea to not stop here, and consider the next option instead. Option 4: Look into using the 'collections.abc' module. Inheriting your class from 'collections.abc.Mapping or 'collections.abc.MutableMapping signals to other users that, for all intents and purposes, your class is a mapping * and can be expected to behave that way. It also provides the methods items, values, get and supports the in keyword and equality checks (==) "for free". You can still cast your object to a dict just as you require, but there would probably be little reason to do so. Because of duck typing, bothering to cast your mapping object to a dict would just be an additional unnecessary step the majority of the time. This answer from me about how to use ABCs might also be helpful. As noted in the comments below: it's worth mentioning that doing this the abc way essentially turns your object class into a dict-like class (assuming you use MutableMapping and not the read-only Mapping base class). Everything you would be able to do with dict, you could do with your own class object. This may be, or may not be, desirable. Also consider looking at the numerical abcs in the numbers module: https://docs.python.org/3/library/numbers.html Since you're also casting your object to an int, it might make more sense to essentially turn your class into a full fledged int so that casting isn't necessary. Option 5: Look into using the dataclasses module (Python 3.7+ only), which includes a convenient asdict() utility method. from dataclasses import dataclass, asdict, field, InitVar @dataclass class Wharrgarbl(object): a: int b: int c: int sum: InitVar[int] # note: InitVar will exclude this from the dict version: InitVar[str] = "old" def __post_init__(self, sum, version): self.sum = 6 # this looks like an OP mistake? self.version = str(version) Now you can do this: >>> asdict(Wharrgarbl(1,2,3,4,"X")) {'a': 1, 'b': 2, 'c': 3} Option 6: Use typing.TypedDict, which has been added in python 3.8. NOTE: option 6 is likely NOT what the OP, or other readers based on the title of this question, are looking for. See additional comments below. class Wharrgarbl(TypedDict): a: str b: str c: str Using this option, the resulting object is a dict (emphasis: the object class will be dict, not Wharrgarbl). There is no reason at all to "cast" it to a dict (unless you are making a copy) because it already is. And since the object is a dict, the initialization signature is identical to that of dict and as such it only accepts keyword arguments or another dictionary. >>> w = Wharrgarbl(a=1,b=2,b=3) >>> w {'a': 1, 'b': 2, 'c': 3} >>> type(w) <class 'dict'> Emphasized: the above "class" Wharrgarbl isn't actually a new class at all. It is simply syntactic sugar for creating typed dict objects with specific keys ONLY and value fields of different types for the type checker. At run time, it is still nothing more than a dict. This option can be pretty convenient for signaling to readers of your code (and also to a type checker such as mypy) that these particular dict objects are expected to have specific keys with specific value types. And it also will allow code completion tools, such as Pycharm or the VSCode python extension, to do their magic. But this means you cannot, for example, add other methods, although you can try: class MyDict(TypedDict): def my_fancy_method(self): return "world changing result" ...but it won't work: >>> MyDict().my_fancy_method() Traceback (most recent call last): File "<stdin>", line 1, in <module> AttributeError: 'dict' object has no attribute 'my_fancy_method' * "Mapping" has become the standard "name" of the dict-like duck type