在 Python 中转换字典的嵌套列表

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

Answer 1

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)

Answer 2

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': {}
#         }
#     }
# }