Converting a nested list of dictionary in Python

Question

I would like to convert a nested list of dictionary into a substructure. And finding a robust way to do so. The structure:

nested_list = [
  {
     "id" : "fruit",
     "name" : "apple"
  },
  {
     "name": "fruit"
  },
  {
     "id" : "fruit",
     "name" : "grape"
  },
  {
     "id" : "fruit",
     "name" : "pineapple"
  },
  {
     "name": "vehicle"
  },
  {
    "id" : "vehicle",
     "name": "car"
  },
  {
    "id" : "car",
     "name": "sedan"
  },
]

Into:

{
  "vehicle": {
    "car": {
        "sedan" : {}
     }
  },
  "fruit" : {
     "apple": {},
    "grape": {},
    "pineapple": {}
  }
}

Note that in this case it can go two level down. But It can go three deep down as well. For example one additional entry:

{ 
   "id" : "sedan", 
   "name": "mini sedan" 
}

My approach so far is:

for category in nested_list:
    if 'id' not in category:
        d[category['name']] = {}

for category in nested_list:
    if 'id' in category and category['id'] in d:
        d[category['id']][category['name']] = {}
    elif 'id' in category and category['id'] not in d:
        for k, v in d.items():
            if category['id'] in v:
                d[k][category['id']] = {category['name']: {}}
    # If there are not top level key then do nothing
    else:
        pass

It works in this case. The problem is it's not robust enough. I'm thinking recursion but unable to crack it. Can someone help out? Thank you

Hi, when you say "nested list", is that intentional? The list you provided is not nested. It contains dictionaries all at the same level. — Alan
– Alan, Commented Mar 30, 2020 at 14:18
Oh Im wrong. It's not nested list. But rather a strange structure with hierarchy — Yehui He
– Yehui He, Commented Mar 30, 2020 at 14:21
problematic input? The first list is the a potential input. I went to have a conversion function to return output like second dictionary — Yehui He
– Yehui He, Commented Mar 30, 2020 at 14:23

Boseong Choi · Accepted Answer · 2020-03-30 14:42:43Z

4

Solution

You can use collections.defaultdict and dict.setdefault:

from collections import defaultdict

nested_list = [
    {
        "id": "fruit",
        "name": "apple"
    },
    {
        "name": "fruit"
    },
    {
        "id": "fruit",
        "name": "grape"
    },
    {
        "id": "fruit",
        "name": "pineapple"
    },
    {
        "name": "vehicle"
    },
    {
        "id": "vehicle",
        "name": "car"
    },
    {
        "id": "car",
        "name": "sedan"
    },
    {
        "id": "sedan",
        "name": "mini sedan"
    },
]

working_dict = defaultdict(dict)
result_dict = {}

for item in nested_list:
    name = item['name']
    if 'id' in item:
        id_ = item['id']
        working_dict[id_].setdefault(name, working_dict[name])
    else:
        result_dict[name] = working_dict[name]
print(working_dict)
print(result_dict)

output:

defaultdict(<class 'dict'>, {'fruit': {'apple': {}, 'grape': {}, 'pineapple': {}}, 'apple': {}, 'grape': {}, 'pineapple': {}, 'vehicle': {'car': {'sedan': {'mini sedan': {}}}}, 'car': {'sedan': {'mini sedan': {}}}, 'sedan': {'mini sedan': {}}, 'mini sedan': {}})
{'fruit': {'apple': {}, 'grape': {}, 'pineapple': {}}, 'vehicle': {'car': {'sedan': {'mini sedan': {}}}}}

Explanation

The idea: dict is mutable.
working_dict is reference table for all "id"s.
If there is no such id, register {} for it.
And register elements without id field, as root elements into result_dict.

Append

If you don't want to use collections.defaultdict, you can only use dict.setdefault. But it is more verbose.

working_dict = {}
result_dict = {}

for item in nested_list:
    name = item['name']
    if 'id' in item:
        id_ = item['id']
        working_dict.setdefault(id_, {}).setdefault(name, working_dict.setdefault(name, {}))
    else:
        result_dict[name] = working_dict.setdefault(name, {})
print(result_dict)

edited Mar 30, 2020 at 14:42

answered Mar 30, 2020 at 14:29

Boseong Choi

2,59611 silver badges24 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Yehui He Over a year ago

this is good. I thought using defaultdict as well. But I want a answer similar to: {'fruit': {'apple': {}, 'grape': {}, 'pineapple': {}}, 'vehicle': {'car': {'sedan': {'mini sedan': {}}}}}

Boseong Choi Over a year ago

result_dict is it.

Yehui He Over a year ago

Oh wow that's amazing. Thank you so much. Just wondering is there a way for regular code this? If not or take too much time that's fine

Boseong Choi Over a year ago

I don't get what regular code means. Can you explain more?

Yehui He Over a year ago

That means not using defaultdict

|

Alexandre B. · Accepted Answer · 2020-03-30 15:05:25Z

You can also do it manually with a recursive function. The idea:

Iterate over the element in the input list
Ignore element without id key
For element with an id in keys:
- recursively search for this key in the out
- add the element (if element already existing, the element is add in the recursive function, else after).

# Recursive search
def iterdictAdd(d, id, name):
    # For all element in the dict
    for k, v in d.items():
        # If key match -> add
        if k == id:
            d[id] = {**d[id], **{name: {}}}
            return True
        else:
            # Recursive call
            if isinstance(v, dict):
                if iterdictAdd(v, id, name):
                    return True
    return False


out = {}
# Iterate all dict
for dict_ in nested_list:
    # Ignore elements without "id" 
    if "id" in dict_:
        # Search and add this key
        if not iterdictAdd(out, dict_["id"], dict_["name"]):
            out[dict_["id"]] = {dict_["name"]: {}}

print(out)
# {'fruit': {
#     'apple': {},
#     'grape': {},
#     'pineapple': {}
# },
#  'vehicle': {
#     'car': {
#         'sedan': {}
#         }
#     }
# }

Collectives™ on Stack Overflow

Converting a nested list of dictionary in Python

2 Answers 2

Solution

Explanation

Append

6 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Solution

Explanation

Append

6 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related