主页 / 问题 / 1212769
Accepted
Inventor
Asked:2021-12-02 15:45:06 +0000 UTC2021-12-02 15:45:06 +0000 UTC

如何将路径转换为字典？Python

  • 772

最平庸的任务，但我不明白如何解决它。如果可能，那么没有第三方库。

最重要的是，我有一个包含路径的列表，例如：

products = [
    "products",
    "products.clothes",
    "products.clothes.jeans",
    "products.clothes.shirts",
    "products.footwear",
    "products.footwear.sneakers",
    "products.footwear.boots",
    "products.accessories"
]

列表中可以有任意数量的元素。路径也可以由任意数量的元素组成。

我需要将整个事情转换成字典并得到这个：

elements = {
    "products": {
        "clothes": {
            "jeans": None,
            "shirts": None
        },
        "footwear": {
            "sneakers": None,
            "boots": None
        },
        "accessories": None
    }
}

我怎样才能做到这一点？

python

3 个回答

  1. Best Answer
    products = [
    "products",
    "products.clothes",
    "products.clothes.jeans",
    "products.clothes.shirts",
    "products.footwear",
    "products.footwear.sneakers",
    "products.footwear.boots",
    "products.accessories"
]

res = dict()

for object in products:
    record = res
    for element in object.split('.'):
        if element not in record:
            record.update({element: None})
        else:
            if record[element] is None:
                record[element] = dict()
            record = record[element]

print(res)

    您可以稍微减少代码（NO :)）：

    res = dict()

for object in products:
    record = res
    for element in object.split('.'):
        record[element] = None if element not in record else dict() if record[element] is None else record[element]
        record = record[element]
    • 5

  2. 也许有人会派上用场，与答案相同，但嵌套可以是任何。

    def create_dict():
    products = [
        "products",
        "products.clothes",
        "products.clothes.jeans",
        "products.clothes.shirts",
        "products.footwear",
        "products.footwear.sneakers",
        "products.footwear.boots",
        "products.accessories",
        "products.accessories.1.2",
        "products.accessories.1.2.3.4.5.clothes",
    ]

    def deep_merge(source, destination):
        """
        :param source - dict to merge
        :param destination - destination dict, will be updated
        :return updated destination
        """
        for key, value in source.items():
            if isinstance(value, dict):
                # получаем или создаём узел n
                n = destination.setdefault(key, {})
                deep_merge(value, n)
            else:
                destination[key] = value
        return destination

    # тут будет результат
    res = dict()
    for e in products:
        # получаем ключи из поля description в обратном порядке
        keys = e.split('.')[::-1]
        # узел будущего словаря
        node = {}
        # флаг который показывает, подставили ли value в конец словаря, елси да то последующие ключи подставляем
        # выше по топологии
        f = True
        # идём по нашим ключам
        # цель, собрать словарь node используя ключи полученные из элемента products
        for k in keys:
            if f:
                # тут вместо None должен быть словарь, для работы рекурсии
                node.update({k: {}})
                f = False
            else:
                node = {k: node}

        # мерджим полученный словарь с основным
        deep_merge(node, res)

    return res

    结果：

    {
  "products": {
    "clothes": {
      "jeans": {},
      "shirts": {}
    },
    "footwear": {
      "sneakers": {},
      "boots": {}
    },
    "accessories": {
      "1": {
        "2": {
          "3": {
            "4": {
              "5": {
                "clothes": {}
              }
            }
          }
        }
      }
    }
  }
}
    • 2 
  3. def setdefault(sline, dt):
    for p in sline[:-1]:
        dt = dt.setdefault(p, {})
    dt.setdefault(sline[-1])

dt = {}
for line in sorted(products, key=lambda x: str.count(x, '.'), reverse=True):
    setdefault(line.split('.'), dt)
print(dt)
    • 0

相关问题