如何从给定嵌套字典生成指定格式的DataFrame？

我有如下结构的嵌套字典：

[
    {
        "name": "A",
        "flag": "folder",
        "children": [
            {
                "name": "A1",
                "flag": "folder",
                "children": [
                    {
                        "name": "A1x",
                        "flag": "file",
                        "children": []
                    },
                    {
                        "name": "A1y",
                        "flag": "file",
                        "children": []
                    }
                ]
            }
        ]
    }
]

希望从中生成指定格式的DataFrame（如附图所示），请问是否有简洁高效的实现方式？

可以通过递归遍历嵌套字典收集节点信息，再转换为DataFrame，这种方式时间复杂度为O(n)（n为节点总数），简洁且高效。以下提供两种常见目标格式的实现：

1. 生成含完整路径与类型的DataFrame #

适合需要直接查看文件/文件夹完整路径的场景：

import pandas as pd

def traverse(node, path=""):
    # 拼接当前节点的完整路径
    current_path = f"{path}/{node['name']}" if path else node['name']
    yield {"路径": current_path, "类型": node['flag']}
    # 递归遍历子节点
    for child in node['children']:
        yield from traverse(child, current_path)

# 原始嵌套字典数据
data = [
    {
        "name": "A",
        "flag": "folder",
        "children": [
            {
                "name": "A1",
                "flag": "folder",
                "children": [
                    {
                        "name": "A1x",
                        "flag": "file",
                        "children": []
                    },
                    {
                        "name": "A1y",
                        "flag": "file",
                        "children": []
                    }
                ]
            }
        ]
    }
]

# 收集所有节点记录
records = []
for root in data:
    records.extend(list(traverse(root)))

# 转换为DataFrame
df = pd.DataFrame(records)
print(df)

输出结果：

路径     类型
0         A  folder
1       A/A1  folder
2    A/A1/A1x    file
3    A/A1/A1y    file

2. 生成层级分列的DataFrame #

适合需要按层级展示目录结构的场景：

import pandas as pd

def traverse_hierarchy(node, levels=None):
    if levels is None:
        levels = []
    # 记录当前节点的层级路径
    current_levels = levels + [node['name']]
    yield {"层级路径": current_levels, "类型": node['flag']}
    # 递归遍历子节点
    for child in node['children']:
        yield from traverse_hierarchy(child, current_levels)

# 收集所有节点的层级信息
records_hierarchy = []
for root in data:
    records_hierarchy.extend(list(traverse_hierarchy(root)))

# 确定最大层级数，补全空值
max_level = max(len(r['层级路径']) for r in records_hierarchy)
# 转换为层级分列的格式
df_hierarchy = pd.DataFrame([
    {f"层级{i+1}": r['层级路径'][i] if i < len(r['层级路径']) else None for i in range(max_level)}
    | {"类型": r['类型']}
    for r in records_hierarchy
])
print(df_hierarchy)

输出结果：

层级1   层级2   层级3     类型
0    A  None  None  folder
1    A    A1  None  folder
2    A    A1   A1x    file
3    A    A1   A1y    file

内容的提问来源于stack exchange，提问作者Pengin