我有字典需要将它填充到树中，如何？

Question

我有这样的字典，需要将它填充到数组或数据库中的树状方案中，例如：

a = {"seasons": "episodes", "peka": {"lol": "wow", "kek": {"wtf": "is this"}}, "ololo": "wololo"} 

键"seasons"拥有自己的ID = 1，Parent_ID = NONE 和值 "episode" 有自己的 ID = 2 和 Parent_ID = 1 ， 和其他字典一样。

Answer 1

/!\ WARNING 不保证字典中的顺序见here

a = { 
      "seasons": "episodes", 
      "peka": {"lol": "wow", "kek": {"wtf": "is this"}}, 
      "ololo": "wololo"
    }

对象a是字典，不保证(key, value)的顺序，也就是说随机如果你做print(a)，你有：

{'ololo': 'wololo', 'peka': {'kek': {'wtf': 'is this'}, 'lol': 'wow'}, 'seasons': 'episodes'}

这是另一个命令。

要保持相同的顺序，请在文件file.json 和用户中复制/粘贴此类型 有序字典。

file.json：

{ 
  "seasons": "episodes", 
  "peka": {"lol": "wow", "kek": {"wtf": "is this"}}, 
  "ololo": "wololo"
}

这里是你的解决方案：

import json
from collections import OrderedDict
from pprint import pprint

with open('file.json', 'r') as filename:
    a = json.load(filename, object_pairs_hook=OrderedDict)


def build_item(_id, parent_id, value):
    return {'ID': _id, 'Parent_ID': parent_id, 'Value': value}


def dfs(_id, root, tree):
    _id += 1
    flat_tree = [build_item(_id, None, root)]
    stack = [(_id, tree)]
    while len(stack) != 0:
        parent_id, tree = stack.pop(0)
        if isinstance(tree, dict):
            for value in tree.keys():
                _id += 1
                flat_tree.append(build_item(_id, parent_id, value))
                stack.append((_id, tree[value]))
        else:
            value = tree
            _id += 1
            flat_tree.append(build_item(_id, parent_id, value))
    return _id, flat_tree


def convert_dict_to_flat_tree(d):
    flat_trees = list()
    _id = 0
    for root, tree in d.items():
        _id, flat_tree = dfs(_id, root, tree)
        flat_trees.extend(flat_tree)
    return flat_trees


flat_tree = convert_dict_to_flat_tree(a)

pprint(flat_tree)

输出：

[{'ID': 1, 'Parent_ID': None, 'Value': 'seasons'},
 {'ID': 2, 'Parent_ID': 1, 'Value': 'episodes'},
 {'ID': 3, 'Parent_ID': None, 'Value': 'peka'},
 {'ID': 4, 'Parent_ID': 3, 'Value': 'lol'},
 {'ID': 5, 'Parent_ID': 3, 'Value': 'kek'},
 {'ID': 6, 'Parent_ID': 4, 'Value': 'wow'},
 {'ID': 7, 'Parent_ID': 5, 'Value': 'wtf'},
 {'ID': 8, 'Parent_ID': 7, 'Value': 'is this'},
 {'ID': 9, 'Parent_ID': None, 'Value': 'ololo'},
 {'ID': 10, 'Parent_ID': 9, 'Value': 'wololo'}]

Answer 2

你想要这样的东西：

a = {"seasons": "episodes", "peka": {"lol": "wow", "kek": {"wtf": "is this"}}, "ololo": "wololo"}

_id = {}

def newid():
    id = _id.setdefault('foo', 0)
    _id['foo'] += 1
    return id

def flat(dic, parent):
    for k,v in dic.items():
        id = newid()
        yield (id, parent, k, v if not isinstance(v, dict) else None)
        if isinstance(v, dict):
           for tup in flat(v, id):
               yield tup

print list(flat(a, newid()))

哪些打印：

[(1, 0, 'seasons', 'episodes'), 
 (2, 0, 'ololo', 'wololo'), 
 (3, 0, 'peka', None), 
 (4, 3, 'kek', None), 
 (5, 4, 'wtf', 'is this'), 
 (6, 3, 'lol', 'wow')]

这些是（ID、Parent ID、Key、Value？）形式的元组。我宁愿输出 E(ID, Parent ID, Key) V(ID, Value)。