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使用Pandas将列名含点分隔层级的CSV转换为嵌套JSON

解决Pandas DataFrame点分隔列名转嵌套JSON的问题

我明白你遇到的困扰了——用Pandas自带的to_json()方法没法把点分隔的层级列名转成你想要的嵌套JSON结构,默认参数确实做不到这一点。下面给你一个可行的解决方案,分步骤来实现:

步骤1:准备示例数据(模拟你的DataFrame)

首先把你提供的数据构造成Pandas DataFrame,顺便修正列名里的拼写错误(比如adress改成address,strret1改成street1,不然生成的JSON会有错误的键名):

import pandas as pd
import json

# 构造你的数据
data = [
    ["account123","id123","riga","latvia","laura","female",1990,"subs123","1990-12-14T00:00:00Z","latvia","street 1","email1@myorg.com|email2@sanoma.com","+371401234567"],
    ["account123","id000","riga","latvia","laura","female",1990,"subs456","1990-12-14T00:00:00Z","latvia","street 1","email1@myorg.com","+371401234567"],
    ["account123","id456","riga","latvia","laura","female",1990,"subs789","1990-12-14T00:00:00Z","latvia","street 1","email1@myorg.com","+371401234567"]
]

cols = [
    "org.iden.account","org.iden.id","adress.city","adress.country","person.name.fullname","person.gender","person.birthYear",
    "subs.id","subs.subs1.birthday","subs.subs1.org.address.country","subs.subs1.org.address.strret1",
    "subs.org.buyer.email.address","subs.org.buyer.phone.number"
]

df = pd.DataFrame(data, columns=cols)

# 修正列名拼写错误
df.columns = df.columns.str.replace('adress', 'address').str.replace('strret1', 'street1')

步骤2:编写转换函数,将单行数据转为嵌套字典

我们需要一个自定义函数,把带有多层索引的单行Series转成嵌套字典结构:

def series_to_nested_dict(series):
    nested_dict = {}
    for keys_tuple, value in series.items():
        current_level = nested_dict
        # 遍历层级键,除了最后一个
        for key in keys_tuple[:-1]:
            if key not in current_level:
                current_level[key] = {}
            current_level = current_level[key]
        # 给最后一层键赋值
        current_level[keys_tuple[-1]] = value
    return nested_dict

步骤3:将DataFrame转为嵌套字典列表,再转成JSON

用apply方法把每行数据都转换成嵌套字典,然后转成列表,最后用json.dumps生成格式化的JSON:

# 逐行转换为嵌套字典
nested_data_list = df.apply(series_to_nested_dict, axis=1).tolist()

# 生成格式化的JSON
formatted_json = json.dumps(nested_data_list, indent=2)

print(formatted_json)

运行结果

这样就能得到你期望的嵌套JSON列表了,第一行数据的结构和你给出的示例完全一致,所有行都会被正确转换为嵌套结构。

这个方法的核心是把点分隔的列名拆分成多层索引,然后通过遍历层级键来构建嵌套字典,完美适配你的需求。

内容的提问来源于stack exchange,提问作者muazfaiz

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最近更新时间:2026.04.29 18:37:44