Creating a recursive python dictionary

is there a way to dynamically create missing keys if I want to set a variable in a subtask.

Essentially I want to create any missing keys and set the value.

self.portdict[switchname][str(neighbor['name'])]['local']['ports'] = []


I am currently doing this, but its a mess:

if not switchname in self.portdict:
    self.portdict[switchname] = {}
if not str(neighbor['name']) in self.portdict[switchname]:
    self.portdict[switchname][str(neighbor['name'])] = {}
if not 'local' in self.portdict[switchname][str(neighbor['name'])]:
    self.portdict[switchname][str(neighbor['name'])]['local'] = {}
if not 'ports' in self.portdict[switchname][str(neighbor['name'])]['local']:
    self.portdict[switchname][str(neighbor['name'])]['local']['ports'] = []


Is there a way to do this in one or two lines?


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4 answers

It's easier to do this without recursion:

def set_by_path(dct, path, value):
    ipath = iter(path)
    p_last = next(ipath)
        while True:
            p_next = next(ipath)
            dct = dct.setdefault(p_last, {})
            p_last = p_next
    except StopIteration:
        dct[p_last] = value


And a test case:

d = {}
set_by_path(d, ['foo', 'bar', 'baz'], 'qux')
print d  # {'foo': {'bar': {'baz': 'qux'}}}


If you want to use it so you don't need a function, you can use the following defaultdict factory, which allows you to nest things deeply:

from collections import defaultdict

defaultdict_factory = lambda : defaultdict(defaultdict_factory)

d = defaultdict_factory()
d['foo']['bar']['baz'] = 'qux'
print d




Use collections.defaultdict

self.portdict = defaultdict(lambda: defaultdict(lambda: defaultdict(lambda: defaultdict(lambda: []))))




I have faced a similar problem in the past. I found that defaultdict

was the correct answer for me, but writing super-long definitions (like @ o11c's answer or @ Apero's answer) didn't help. Here's what I came up with instead:

from collections import defaultdict
from functools import partial

def NestedDefaultDict(levels, baseFn):
    def NDD(lvl):
        return partial(defaultdict, NDD(lvl-1)) if lvl > 0 else baseFn
    return defaultdict(NDD(levels-1))


This creates a dictionary with levels

nested dictionaries. So if you have levels

= 3, you need 3 keys to access the lower level value. The second argument is the function that is used to create the lower level values. Something like list

or lambda: 0

or even dict

would work well.

Here's an example of using "automatic" keys with 4 levels

and list

as the default function:

>>> x = NestedDefaultDict(4, list)
>>> x[1][2][3][4].append('hello')
>>> x
defaultdict(<functools.partial object at 0x10b5c22b8>, {1: defaultdict(<functools.partial object at 0x10b5c2260>, {2: defaultdict(<functools.partial object at 0x10b5c2208>, {3: defaultdict(<type 'list'>, {4: ['hello']})})})})


I think that's basically what you want for the case in your question. Your 4 "layers" are switchname, neighborname, localhost and ports, and it looks like you want list

the lower layer to keep your ports.

Another example: 2 levels

and lambda: 0

by default:

>>> y = NestedDefaultDict(2, lambda: 0)
>>> y['foo']['bar'] += 7
>>> y['foo']['baz'] += 10
>>> y['foo']['bar'] += 1
>>> y
defaultdict(<functools.partial object at 0x1021f1310>, {'foo': defaultdict(<function <lambda> at 0x1021f3938>, {'baz': 10, 'bar': 8})})




Take a close look at collection.defaultdict:

from collections import defaultdict
foo = defaultdict(dict)
foo['bar'] = defaultdict(dict)
foo['bar']['baz'] = defaultdict(dict)
foo['bar']['baz']['aaa'] = 1
foo['bor'] = 0
foo['bir'] = defaultdict(list)

print foo
defaultdict(<type 'dict'>, {'bir': defaultdict(<type 'list'>, {'biz': [1, 2]}), 'bor': 0, 'bar': defaultdict(<type 'dict'>, {'baz': defaultdict(<type 'dict'>, {'aaa': 1})})})




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