How to change behavior of dict() for an instance

Question

So I'm writing a class that extends a dictionary which right now uses a method "dictify" to transform itself into a dict. What I would like to do instead though is change it so that calling dict() on the object results in the same behavior, but I don't know which method to override. Is this not possible, or I am I missing something totally obvious? (And yes, I know the code below doesn't work but I hope it illustrates what I'm trying to do.)

from collections import defaultdict

class RecursiveDict(defaultdict):
    '''
    A recursive default dict.

    >>> a = RecursiveDict()
    >>> a[1][2][3] = 4
    >>> a.dictify()
    {1: {2: {3: 4}}}
    '''
    def __init__(self):
        super(RecursiveDict, self).__init__(RecursiveDict)

    def dictify(self):
        '''Get a standard dictionary of the items in the tree.'''
        return dict([(k, (v.dictify() if isinstance(v, dict) else v))
                     for (k, v) in self.items()])

    def __dict__(self):
        '''Get a standard dictionary of the items in the tree.'''
        print [(k, v) for (k, v) in self.items()]
        return dict([(k, (dict(v) if isinstance(v, dict) else v))
                     for (k, v) in self.items()])

EDIT: To show the problem more clearly:

>>> b = RecursiveDict()
>>> b[1][2][3] = 4
>>> b
defaultdict(<class '__main__.RecursiveDict'>, {1: defaultdict(<class '__main__.RecursiveDict'>, {2: defaultdict(<class '__main__.RecursiveDict'>, {3: 4})})})
>>> dict(b)
{1: defaultdict(<class '__main__.RecursiveDict'>, {2: defaultdict(<class '__main__.RecursiveDict'>, {3: 4})})}
>>> b.dictify()
{1: {2: {3: 4}}}

I want dict(b) to be same as b.dictify()

Answer 1

Nothing wrong with your approach, but this is similar to the Autovivification feature of Perl, which has been implemented in Python in this question. Props to @nosklo for this.

class RecursiveDict(dict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value

>>> a = RecursiveDict()
>>> a[1][2][3] = 4
>>> dict(a)
{1: {2: {3: 4}}}

EDIT

As suggested by @Rosh Oxymoron, using __missing__ results in a more concise implementation. Requires Python >= 2.5

class RecursiveDict(dict):
    """Implementation of perl's autovivification feature."""
    def __missing__(self, key):
        value = self[key] = type(self)()
        return value

Answer 2

Do you want just to print it like a dict ? use this:

from collections import defaultdict

class RecursiveDict(defaultdict):
    '''
    A recursive default dict.

    >>> a = RecursiveDict()
    >>> a[1][2][3] = 4
    >>> a.dictify()
    {1: {2: {3: 4}}}
    >>> dict(a)
    {1: {2: {3: 4}}}

    '''
    def __init__(self):
        super(RecursiveDict, self).__init__(RecursiveDict)

    def dictify(self):
        '''Get a standard dictionary of the items in the tree.'''
        return dict([(k, (v.dictify() if isinstance(v, dict) else v))
                     for (k, v) in self.items()])

    def __dict__(self):
        '''Get a standard dictionary of the items in the tree.'''
        print [(k, v) for (k, v) in self.items()]
        return dict([(k, (dict(v) if isinstance(v, dict) else v))
                     for (k, v) in self.items()])

    def __repr__(self):
        return repr(self.dictify())

Maybe you are looking for __missing__ :

class RecursiveDict(dict):
    '''
    A recursive default dict.

    >>> a = RecursiveDict()
    >>> a[1][2][3] = 4
    >>> a
    {1: {2: {3: 4}}}
    >>> dict(a)
    {1: {2: {3: 4}}}

    '''

    def __missing__(self, key):
        self[key] = self.__class__()
        return self[key]

Answer 3

edit: As ironchefpython pointed out in comments, this isn't actually doing what I thought it did, as in my example b[1] is still a RecursiveDict. This may still be useful, as you essentially get an object pretty similar Rob Cowie's answer, but it is built on defaultdict.

---

You can get the behavior you want (or something very similar) by overriding __repr__, check this out:

class RecursiveDict(defaultdict):
    def __init__(self):
        super(RecursiveDict, self).__init__(RecursiveDict)

    def __repr__(self):
        return repr(dict(self))

>>> a = RecursiveDict()
>>> a[1][2][3] = 4
>>> a             # a looks like a normal dict since repr is overridden
{1: {2: {3: 4}}}
>>> type(a)
<class '__main__.RecursiveDict'>
>>> b = dict(a)
>>> b             # dict(a) gives us a normal dictionary
{1: {2: {3: 4}}}
>>> b[5][6] = 7   # obviously this won't work anymore
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 5
>>> type(b)
<type 'dict'>

There may be a better way to get to a normal dictionary view of the defaultdict than dict(self) but I couldn't find one, comment if you know how.

Answer 4

You can't do it.

I deleted my previous answer, because I found after looking at the source code, that if you call dict(d) on a d that is a subclass of dict, it makes a fast copy of the underlying hash in C, and returns a new dict object.

Sorry.

If you really want this behavior, you'll need to create a RecursiveDict class that doesn't inherit from dict, and implement the __iter__ interface.