Checking a nested dictionary using a dot notation string “a.b.c.d.e”, automatically create missing levels

问题

This one is blowing my mind. Given the following dictionary:

    d = {\"a\":{\"b\":{\"c\":\"winning!\"}}}

I have this string (from an external source, and I can\'t change this metaphor).

    k = \"a.b.c\"

I need to determine if the dictionary has the key \'c\', so I can add it if it doesn\'t.

This works swimmingly for retrieving a dot notation value:

    reduce(dict.get, key.split(\".\"), d)

but I can\'t figure out how to \'reduce\' a has_key check or anything like that.

My ultimate problem is this: given \"a.b.c.d.e\", I need to create all the elements necessary in the dictionary, but not stomp them if they already exist. If someone knows a whizbang way to do all that, you\'ll be my hero.

回答1:

... or using recursion:

def put(d, keys, item):
    if "." in keys:
        key, rest = keys.split(".", 1)
        if key not in d:
            d[key] = {}
        put(d[key], rest, item)
    else:
        d[keys] = item

def get(d, keys):
    if "." in keys:
        key, rest = keys.split(".", 1)
        return get(d[key], rest)
    else:
        return d[keys]

回答2:

You could use an infinite, nested defaultdict:

>>> from collections import defaultdict
>>> infinitedict = lambda: defaultdict(infinitedict)
>>> d = infinitedict()
>>> d['key1']['key2']['key3']['key4']['key5'] = 'test'
>>> d['key1']['key2']['key3']['key4']['key5']
'test'

Given your dotted string, here's what you can do:

>>> import operator
>>> keys = "a.b.c".split(".")
>>> lastplace = reduce(operator.getitem, keys[:-1], d)
>>> lastplace.has_key(keys[-1])
False

You can set a value:

>>> lastplace[keys[-1]] = "something"
>>> reduce(operator.getitem, keys, d)
'something'
>>> d['a']['b']['c']
'something'

回答3:

How about an iterative approach?

def create_keys(d, keys):
    for k in keys.split("."):
        if not k in d: d[k] = {}  #if the key isn't there yet add it to d
        d = d[k]                  #go one level down and repeat

If you need the last key value to map to anything else than a dictionary you could pass the value as an additional argument and set this after the loop:

def create_keys(d, keys, value):
    keys = keys.split(".")
    for k in keys[:-1]:
        if not k in d: d[k] = {}
        d = d[k]            
    d[keys[-1]] = value

回答4:

d = {"a":{}}
k = "a.b.c".split(".")

def f(d, i):
    if i >= len(k):
        return "winning!"
    c = k[i]
    d[c] = f(d.get(c, {}), i + 1)
    return d

print f(d, 0)
"{'a': {'b': {'c': 'winning!'}}}"

来源：https://stackoverflow.com/questions/12414821/checking-a-nested-dictionary-using-a-dot-notation-string-a-b-c-d-e-automatica

标签

python

dictionary

nested

defaultdict