提取在嵌套字典和列表中找到的叶值集，不包括 None

Question

我有一个从 YAML 中读取的嵌套结构，它由嵌套列表 and/or 嵌套字典或在不同嵌套级别的混合组成。可以假定该结构不包含任何递归对象。

如何只从中提取叶值？另外，我不想要任何 None 值。叶值包含字符串，这是我所关心的。考虑到结构的最大深度不足以超过堆栈递归限制，使用递归是可以的。发电机也可以选择。

存在处理扁平化列表或字典的类似问题，但不是两者的混合。或者，如果扁平化字典，它们也会 return 我并不真正需要的扁平化键，并且有名称冲突的风险。

我试过 more_itertools.collapse 但它的示例仅显示它适用于嵌套列表，而不适用于字典和列表的混合。

示例输入

struct1 = {
    "k0": None,
    "k1": "v1",
    "k2": ["v0", None, "v1"],
    "k3": ["v0", ["v1", "v2", None, ["v3"], ["v4", "v5"], []]],
    "k4": {"k0": None},
    "k5": {"k1": {"k2": {"k3": "v3", "k4": "v6"}, "k4": {}}},
    "k6": [{}, {"k1": "v7"}, {"k2": "v8", "k3": "v9", "k4": {"k5": {"k6": "v10"}, "k7": {}}}],
    "k7": {
        "k0": [],
        "k1": ["v11"],
        "k2": ["v12", "v13"],
        "k3": ["v14", ["v15"]],
        "k4": [["v16"], ["v17"]],
        "k5": ["v18", ["v19", "v20", ["v21", "v22", []]]],
    },
}

struct2 = ["aa", "bb", "cc", ["dd", "ee", ["ff", "gg"], None, []]]

预期产出

struct1_leaves = {f"v{i}" for i in range(23)}
struct2_leaves = {f"{s}{s}" for s in "abcdefg"}

Answer 1

这是一个简单的参考解决方案，它使用递归为问题中包含的示例输入生成预期输出。

from typing import Any, Set


def leaves(struct: Any) -> Set[Any]:
    """Return a set of leaf values found in nested dicts and lists excluding None values."""
    # Ref: 
    values = set()
    if isinstance(struct, dict):
        for sub_struct in struct.values():
            values.update(leaves(sub_struct))
    elif isinstance(struct, list):
        for sub_struct in struct:
            values.update(leaves(sub_struct))
    elif struct is not None:
        values.add(struct)
    return values

Answer 2

另一种可能性是使用带递归的生成器：

struct1 = {'k0': None, 'k1': 'v1', 'k2': ['v0', None, 'v1'], 'k3': ['v0', ['v1', 'v2', None, ['v3'], ['v4', 'v5'], []]], 'k4': {'k0': None}, 'k5': {'k1': {'k2': {'k3': 'v3', 'k4': 'v6'}, 'k4': {}}}, 'k6': [{}, {'k1': 'v7'}, {'k2': 'v8', 'k3': 'v9', 'k4': {'k5': {'k6': 'v10'}, 'k7': {}}}], 'k7': {'k0': [], 'k1': ['v11'], 'k2': ['v12', 'v13'], 'k3': ['v14', ['v15']], 'k4': [['v16'], ['v17']], 'k5': ['v18', ['v19', 'v20', ['v21', 'v22', []]]]}}
def flatten(d):
   for i in getattr(d, 'values', lambda :d)():
      if isinstance(i, str):
         yield i
      elif i is not None:
         yield from flatten(i)

print(set(flatten(struct1)))

输出：

{'v10', 'v9', 'v8', 'v7', 'v0', 'v18', 'v16', 'v1', 'v21', 'v11', 'v14', 'v15', 'v12', 'v13', 'v4', 'v2', 'v5', 'v20', 'v6', 'v19', 'v3', 'v22', 'v17'}

struct2 = ["aa", "bb", "cc", ["dd", "ee", ["ff", "gg"], None, []]]
print(set(flatten(struct2)))

输出：

{'cc', 'ff', 'dd', 'gg', 'bb', 'ee', 'aa'}

Answer 3

这是对的改编，使用内部函数和单个 set。它还使用递归为问题中包含的样本输入生成预期输出。它避免了通过整个调用堆栈传递每个叶子。

from typing import Any, Set


def leaves(struct: Any) -> Set[Any]:
    """Return a set of leaf values found in nested dicts and lists excluding None values."""
    # Ref: 
    values = set()

    def add_leaves(struct_: Any) -> None:
        if isinstance(struct_, dict):
            for sub_struct in struct_.values():
                add_leaves(sub_struct)
        elif isinstance(struct_, list):
            for sub_struct in struct_:
                add_leaves(sub_struct)
        elif struct_ is not None:
            values.add(struct_)

    add_leaves(struct)
    return values

提取在嵌套字典和列表中找到的叶值集，不包括 None

Extract set of leaf values found in nested dicts and lists excluding None

python

recursion

nested

flatten

示例输入

预期产出