如何将DataFrame中含字典列表的Series转换为指定格式DataFrame？

安全阿岳

2026-7-11

解决方法

方法一：直接用`apply`处理`detail`列

遍历detail列的每个字典列表，提取对应字段后用分隔符拼接，直接生成目标列，代码简洁直观：

import pandas as pd

data = [
    {
        "id": 1,
        "detail": [
            {"name": "name1", "type": "type1"},
            {"name": "name2", "type": "type2"},
        ]
    },
    {
        "id": 2,
        "detail": [
            {"name": "name3", "type": "type3"},
            {"name": "name4", "type": "type4"},
        ]
    },
]
df = pd.DataFrame(data)

# 生成拼接后的name和type列
df['detail_name'] = df['detail'].apply(lambda x: ' | '.join([d['name'] for d in x]))
df['detail_type'] = df['detail'].apply(lambda x: ' | '.join([d['type'] for d in x]))

# 保留需要的列
result_df = df[['id', 'detail_name', 'detail_type']]
print(result_df)

输出结果：

id    detail_name    detail_type
0   1  name1 | name2  type1 | type2
1   2  name3 | name4  type3 | type4

方法二：先展开再分组聚合

基于你尝试的pd.json_normalize，先展开所有字典条目，再按id分组拼接，适合需要先对明细做额外处理的场景：

import pandas as pd

data = [
    {
        "id": 1,
        "detail": [
            {"name": "name1", "type": "type1"},
            {"name": "name2", "type": "type2"},
        ]
    },
    {
        "id": 2,
        "detail": [
            {"name": "name3", "type": "type3"},
            {"name": "name4", "type": "type4"},
        ]
    },
]

# 展开detail列表，同时关联原id
details = pd.json_normalize(data, record_path=['detail'], meta=['id'], record_prefix='detail_')

# 按id分组，拼接同组的name和type
result_df = details.groupby('id').agg(
    detail_name=('detail_name', ' | '.join),
    detail_type=('detail_type', ' | '.join)
).reset_index()

print(result_df)

输出结果与方法一完全一致。