Nested dictionary value from key path

Question

Get the value from a nested dictionary with the help of key path, here is the dict:

json = {
    "app": {
        "Garden": {
            "Flowers": {
                "Red flower": "Rose",
                "White Flower": "Jasmine",
                "Yellow Flower": "Marigold"
            }
        },
        "Fruits": {
            "Yellow fruit": "Mango",
            "Green fruit": "Guava",
            "White Flower": "groovy"
        },
        "Trees": {
            "label": {
                "Yellow fruit": "Pumpkin",
                "White Flower": "Bogan"
            }
        }
    }

The input parameter to the method is the key path with dots separated, from the key path = "app.Garden.Flowers.white Flower" need to print 'Jasmine'. My code so far:

import json
with open('data.json') as data_file:    
  j = json.load(data_file)


def find(element, JSON):     
  paths = element.split(".")  
  # print JSON[paths[0]][paths[1]][paths[2]][paths[3]]
  for i in range(0,len(paths)):
    data = JSON[paths[i]]
    # data = data[paths[i+1]]
    print data



find('app.Garden.Flowers.White Flower',j)

score 49 · Answer 1 · edited Jan 10 '22 at 10:53

49

This is an instance of a fold. You can either write it concisely like this:

from functools import reduce
import operator

def find(element, json):
    return reduce(operator.getitem, element.split('.'), json)

Or more Pythonically (because reduce() is frowned upon due to poor readability) like this:

def find(element, json):
    keys = element.split('.')
    rv = json
    for key in keys:
        rv = rv[key]
    return rv

j = {"app": {
    "Garden": {
        "Flowers": {
            "Red flower": "Rose",
            "White Flower": "Jasmine",
            "Yellow Flower": "Marigold"
        }
    },
    "Fruits": {
        "Yellow fruit": "Mango",
        "Green fruit": "Guava",
        "White Flower": "groovy"
    },
    "Trees": {
        "label": {
            "Yellow fruit": "Pumpkin",
            "White Flower": "Bogan"
        }
    }
}}
print find('app.Garden.Flowers.White Flower', j)

edited Jan 10 '22 at 10:53

Tomerikoo

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answered Jun 24 '15 at 17:57

Ceasar Bautista

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1

Instead of defining your own item-getter function using lambda, you can `import operator` and use `operator.getitem`. – Christian Long Oct 31 '17 at 21:37
1

If you want to address an item by an integer index (app.Garden.Flowers.0), you may change your for-loop like this: try: rv = rv[int(key)] except: rv = rv[key] – n.r. 4 mins ago Edit – n.r. Apr 21 '21 at 19:58
1

this solution doesnt support list ,If you change Flowers to be list - u get an error – AviC Jan 10 '22 at 10:41

score 8 · Answer 2 · edited Jan 10 '22 at 09:41

8

I was in a similar situation and found this dpath module. Nice and easy.

edited Jan 10 '22 at 09:41

Tomerikoo

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answered Feb 01 '17 at 12:17

juanra

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Very cool. What I was hoping for, however, is that there's a way to do something like `s = "%('red/buggy/bumpers')s" % { "red": { "buggy": { "bumpers":"foo" }}}`. But I suppose it's better to use a proper templating language than to change the behaviour of the `%` operator on a string. – Michael Scheper Oct 15 '18 at 19:43
I mean you could determine that the substring starting with '{' and ending with '}' is a dictionary. The run that substring through the `eval()` method which turns it into a dictionary object. Then split your '/' delimited path and apply the keys using a for loop – M. Choy Feb 12 '21 at 03:38

Anthon · Answer 3 · 2015-06-24T18:23:28.367

Your code heavily depends on no dots every occurring in the key names, which you might be able to control, but not necessarily.

I would go for a generic solution using a list of element names and then generate the list e.g. by splitting a dotted list of key names:

class ExtendedDict(dict):
    """changes a normal dict into one where you can hand a list
    as first argument to .get() and it will do a recursive lookup
    result = x.get(['a', 'b', 'c'], default_val)
    """
    def multi_level_get(self, key, default=None):
        if not isinstance(key, list):
            return self.get(key, default)
        # assume that the key is a list of recursively accessible dicts
        def get_one_level(key_list, level, d):
            if level >= len(key_list):
                if level > len(key_list):
                    raise IndexError
                return d[key_list[level-1]]
            return get_one_level(key_list, level+1, d[key_list[level-1]])

        try:
            return get_one_level(key, 1, self)
        except KeyError:
            return default

    get = multi_level_get # if you delete this, you can still use the multi_level-get

Once you have this class it is easy to just transform your dict and get "Jasmine":

json = {
        "app": {
            "Garden": {
                "Flowers": {
                    "Red flower": "Rose",
                    "White Flower": "Jasmine",
                    "Yellow Flower": "Marigold"
                }
            },
            "Fruits": {
                "Yellow fruit": "Mango",
                "Green fruit": "Guava",
                "White Flower": "groovy"
            },
            "Trees": {
                "label": {
                    "Yellow fruit": "Pumpkin",
                    "White Flower": "Bogan"
                }
            }
        }
    }

j = ExtendedDict(json)
print j.get('app.Garden.Flowers.White Flower'.split('.'))

will get you:

Jasmine

Like with a normal get() from a dict, you get None if the key (list) you specified doesn't exists anywhere in the tree, and you can specify a second parameter as return value instead of None

score 3 · Answer 4 · edited Oct 29 '20 at 17:27

3

I suggest you to use python-benedict, a python dict subclass with full keypath support and many utility methods.

You just need to cast your existing dict:

d = benedict(json)
# now your keys support dotted keypaths
print(d['app.Garden.Flower.White Flower'])

Here the library and the documentation: https://github.com/fabiocaccamo/python-benedict

Note: I am the author of this project

edited Oct 29 '20 at 17:27

Martijn Pieters

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answered May 20 '19 at 22:13

Fabio Caccamo

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score 2 · Accepted Answer · answered Jun 24 '15 at 17:54

Very close. You need to (as you had in your comment) recursively go through the main JSON object. You can accomplish that by storing the result of the outermost key/value, then using that to get the next key/value, etc. till you're out of paths.

def find(element, JSON):     
  paths = element.split(".")
  data = JSON
  for i in range(0,len(paths)):
    data = data[paths[i]]
  print data

You still need to watch out for KeyErrors though.

score 2 · Answer 6 · answered Mar 20 '21 at 13:40

2

one-liner:

from functools import reduce

a = {"foo" : { "bar" : "blah" }}
path = "foo.bar"

reduce(lambda acc,i: acc[i], path.split('.'), a)

answered Mar 20 '21 at 13:40

Wildhammer

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Yogesh Kamat · Answer 7 · 2021-03-23T19:05:01.443

Option 1: pyats library from Cisco [its a c extension]

Its quick and Super fast (measure it with timeit if required)
Javascript-ish usage [Bracket lookup ,dotted lookup, combined lookup]
Dotted Lookup for missing key raises Attribute error, bracket or default python dict lookup gives KeyError.

pip install pyats pyats-datastructures pyats-utils

from pyats.datastructures import NestedAttrDict
item = {"specifications": {"os": {"value": "Android"}}}
path = "specifications.os.value"
x = NestedAttrDict(item)
print(x[path])# prints Android
print(x['specifications'].os.value)# prints Android
print(x['specifications']['os']['value'])#prints Android
print(x['specifications'].os.value1)# raises Attribute Error

Option 2:pyats.utils chainget

super fast (measure it with timeit if required)

from pyats.utils import utils
item = {"specifications": {"os": {"value": "Android"}}}
path = "specifications.os.value"
path1 = "specifications.os.value1"
print(utils.chainget(item,path))# prints android (string version)
print(utils.chainget(item,path.split('.')))# prints android(array version)
print(utils.chainget(item,path1))# raises KeyError

Option 3: python without external library

Better speed in comparison to lambda.
Separate Error handling not required as in lambda and other cases.
Readable and concise can be a utils function/helper in the project

from functools import reduce
item = {"specifications": {"os": {"value": "Android"}}}
path1 = "specifications.family.value"
path2 = "specifications.family.value1"

def test1():
    print(reduce(dict.get, path1.split('.'), item))

def test2():
    print(reduce(dict.get, path2.split('.'), item))

test1() # prints Android
test2() # prints None

Nested dictionary value from key path

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