tponthieux tponthieux - 1 year ago 86
Python Question

Recursive DotDict

I have a utility class that makes Python dictionaries behave somewhat like JavaScript objects as far as getting and setting attributes.

class DotDict(dict):
"""
a dictionary that supports dot notation
as well as dictionary access notation
usage: d = DotDict() or d = DotDict({'val1':'first'})
set attributes: d.val2 = 'second' or d['val2'] = 'second'
get attributes: d.val2 or d['val2']
"""
__getattr__ = dict.__getitem__
__setattr__ = dict.__setitem__
__delattr__ = dict.__delitem__


I would like to make it so it also converts nested dictionaries into DotDict() instances. I was hoping to be able to do something like this with
__init__
or
__new__
, but I haven't come up with anything that works:

def __init__(self, dct):
for key in dct.keys():
if hasattr(dct[key], 'keys'):
dct[key] = DotDict(dct[key])


How can I recursively convert the nested dictionaries into DotDict() instances?

>>> dct = {'scalar_value':1, 'nested_dict':{'value':2}}
>>> dct = DotDict(dct)

>>> print dct
{'scalar_value': 1, 'nested_dict': {'value': 2}}

>>> print type(dct)
<class '__main__.DotDict'>

>>> print type(dct['nested_dict'])
<type 'dict'>

Answer Source

I don't see where you are copying the values in the constructor. Here DotDict is always empty because of that. When I added the key assignment, it worked:

class DotDict(dict):
    """
    a dictionary that supports dot notation 
    as well as dictionary access notation 
    usage: d = DotDict() or d = DotDict({'val1':'first'})
    set attributes: d.val2 = 'second' or d['val2'] = 'second'
    get attributes: d.val2 or d['val2']
    """
    __getattr__ = dict.__getitem__
    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__

    def __init__(self, dct):
        for key, value in dct.items():
            if hasattr(value, 'keys'):
                value = DotDict(value)
            self[key] = value


dct = {'scalar_value':1, 'nested_dict':{'value':2, 'nested_nested': {'x': 21}}}
dct = DotDict(dct)

print dct.nested_dict.nested_nested.x

It looks a bit dangerous and error prone, not to mention source of countless surprises to other developers, but seems to be working.

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