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73 lines
1.9 KiB
Python
73 lines
1.9 KiB
Python
"""
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Есть абстрактная задача:
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У нас есть лес из json массива
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Гораздо дешевле в качестве транспорта их передавать в виде списка кортежей
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(вида ключ: значение),
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(ключ.ключ: значение)
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Пример:
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"""
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import json
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from typing import Any, List, Tuple
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forest = {'b': 1,
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'c': {'d': 2,
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'f': 3,
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},
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'e': {'i': 4,
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'j': {'k': 5,
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'l': 6,
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},
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},
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'm': {'n': 7,
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'o': {'p': 8,
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'r': {'s': 9,
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't': 10,
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},
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},
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},
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}
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# Переводим в такой вид:
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result = [('b', 1), ('c.d', 2), ('c.f', 3), ('e.i', 4), ('e.j.k', 5), ('e.j.l', 6), ('m.n', 7),
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('m.o.p', 8), ('m.o.r.s', 9), ('m.o.r.t', 10), ]
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def dict_to_list(sub_tree: dict, current_name: str, items_list: List[tuple]) -> List[tuple]:
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for key in sub_tree:
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if isinstance(sub_tree[key], dict):
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dict_to_list(sub_tree=sub_tree[key], current_name=current_name + key, items_list=items_list)
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else:
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items_list.append(('.'.join(current_name + key), sub_tree[key]))
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return items_list
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res = dict_to_list(forest, '', [])
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print(res)
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# Reverse task:
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def list_to_dict(data: List[Tuple['str', int]]) -> dict:
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tree = {}
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for item in data:
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curr_tree = tree
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keys = item[0].split('.')
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for i in range(0, len(keys)):
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if i == len(keys) - 1:
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curr_tree[keys[i]] = item[1]
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elif keys[i] not in curr_tree:
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curr_tree[keys[i]] = {}
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curr_tree = curr_tree[keys[i]]
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return tree
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parsed = json.dumps(list_to_dict(result), indent=4)
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print(parsed)
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