Nested insertion/creation of dictionary

Question

Given that you have an empty dictionary

data = {}

I have a path and a value

path = "root.sub.item"
value = 12

How could I recursively add objects that do not exist?

def add_value(path, value):
    for part in path.split('.'):
        if not part in data:
            data[part] = {}

The expected output for this would be:

data = {
    'root':{
        'sub':{
            'item': 12
        }
    }
}

Could somebody help out with this or point me in the right direction?
I'm using Python 3.6.

score 0 · Answer 1 · answered Apr 17 '18 at 14:13

0

You're almost there, you just need to keep track of how far you are into the tree structure, and a way to know when you're on the last element of the path:

def add_value(path, value):
    tmp = data
    parts = list(path.split('.'))
    for i in range(len(parts) - 1):
        part = parts[i]
        if not part in tmp:
            tmp[part] = {}
        tmp = tmp[part]
    tmp[parts[-1]] = value

answered Apr 17 '18 at 14:13

Isaac

3,586
1
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But this will only give me the last insersion. Eg. {'item': 12} ? – Dennis Delin Apr 17 '18 at 14:25
@DennisDelin, did you run it? When I run it I get `data` with the nested structure you describe. – Isaac Apr 17 '18 at 14:55
Yes if I do a print of tmp at the end of the function it will only hold last dict with the value 12 – Dennis Delin Apr 18 '18 at 11:45
Right, because `tmp` is a temporary variable, which is used to get the values into `data`. What happens if you print `data`? – Isaac Apr 19 '18 at 20:09

JOHN_16 · Accepted Answer · 2018-04-17T16:13:07.180

You can use some another kind of solution like recursive defaultdict, as in this answer.

A quick and stupid example about how it can used:

from collections import defaultdict

def func(rdict, path, value):
    items = path.split('.')

    d = rdict[items[0]]

    for item in items[1:-1]:
        d = d[item]

    d[items[-1]] = value

nested_dict = lambda: defaultdict(nested_dict)

result = nested_dict()

func(result, 'root.sub.item', 12)
func(result, 'root.moon.value', 1)

assert result['root']['sub']['item'] == 12
assert result['root']['moon']['value'] == 1
assert result['root']['moon']['noop'] != 0

score 0 · Answer 3 · answered Apr 18 '18 at 04:30

0

you can try Raymond Hettinger recipe :

source: https://twitter.com/raymondh/status/343823801278140417

from collections import defaultdict

infinity_dict=lambda:defaultdict(infinity_dict)

d=infinity_dict()
d['root']['sub']['item'] = 12

answered Apr 18 '18 at 04:30

Aaditya Ura

12,007
7
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Nested insertion/creation of dictionary

3 Answers3