如何在Python中将复杂嵌套字典转换为Pandas DataFrame

我有一个结构如下的嵌套字典：

dict = {
    'series_1': ['id_series', [
        ['season_1', [['ep1_title', 'ep_url'], ['ep2_title', 'ep_url']...]],
        ['season_2', [['ep1_title', 'ep_url'], ['ep2_title', 'ep_url']...]],
        ...
    ]],
    'series_2': ['id_series', [
        ['season_1', [['ep1_title', 'ep_url'], ['ep2_title', 'ep_url']...]],
        ['season_2', [['ep1_title', 'ep_url'], ['ep2_title', 'ep_url']...]],
        ...
    ]],
    ...
}

示例字典：

{
    'Scooby-Doo! Mystery Incorporated': ['1660055', [
        ['season 1', [['Pawn of Shadows', 'https://dl.opensubtitles.org/it/download/sub/4797725'], ['All Fear the Freak', 'https://dl.opensubtitles.org/it/download/sub/4797755']]],
        ['season 2', [['Through the Curtain', 'https://dl.opensubtitles.org/it/download/sub/5465599'], ['Come Undone', 'https://dl.opensubtitles.org/it/download/sub/5465681']]]
    ]],
    'Scooby e Scrappy Doo': ['0084970', [
        ['season 1', [["Scooby Roo/Scooby's Gold Medal Gambit", 'https://dl.opensubtitles.org/it/download/sub/6086643'], ['The Mark of Scooby/The Crazy Carnival Caper', 'https://dl.opensubtitles.org/it/download/sub/6086649']]]
    ]]
}

希望将其转换为如下格式的Pandas DataFrame：

尝试过其他嵌套字典转DataFrame的方案，但不符合需求，求解决方法。

可以通过遍历嵌套字典的层级结构，收集每一行需要的数据，最后转换为DataFrame。具体代码如下：

import pandas as pd

# 示例字典
data = {
    'Scooby-Doo! Mystery Incorporated': ['1660055', [
        ['season 1', [['Pawn of Shadows', 'https://dl.opensubtitles.org/it/download/sub/4797725'], ['All Fear the Freak', 'https://dl.opensubtitles.org/it/download/sub/4797755']]],
        ['season 2', [['Through the Curtain', 'https://dl.opensubtitles.org/it/download/sub/5465599'], ['Come Undone', 'https://dl.opensubtitles.org/it/download/sub/5465681']]]
    ]],
    'Scooby e Scrappy Doo': ['0084970', [
        ['season 1', [["Scooby Roo/Scooby's Gold Medal Gambit", 'https://dl.opensubtitles.org/it/download/sub/6086643'], ['The Mark of Scooby/The Crazy Carnival Caper', 'https://dl.opensubtitles.org/it/download/sub/6086649']]]
    ]]
}

# 初始化空列表存储每行数据
rows = []

# 遍历每部剧集
for series_title, series_info in data.items():
    id_series = series_info[0]
    seasons = series_info[1]
    
    # 遍历每季
    for season in seasons:
        season_name = season[0]
        # 提取季数（比如从"season 1"中取出1）
        season_number = int(season_name.split()[1])
        episodes = season[1]
        
        # 遍历每集
        for ep in episodes:
            ep_title, ep_url = ep
            # 将数据添加到列表
            rows.append({
                'series_title': series_title,
                'id_series': id_series,
                '#season': season_number,
                'ep_title': ep_title,
                'ep_url': ep_url
            })

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

代码逻辑说明： #

1. 层级遍历：从最外层的剧集名称开始，依次进入剧集ID、季列表、每集信息；
2. 数据提取：从季名称中提取数字作为#season字段；
3. 行数据收集：将每一集对应的所有字段封装为字典，存入列表；
4. 转换DataFrame：用Pandas将列表直接转换为目标格式的DataFrame。

运行代码后，输出的DataFrame就会完全符合你的需求。

内容的提问来源于stack exchange，提问作者Enrico Caria