我知道这个问题已经被问到很长时间了，但我想我可以在其他人的答案上加上一篇博客文章，以一种非常简洁的方式解释解决方案。

这是链接

打开文件进行写入

employ_data = open('/tmp/EmployData.csv', 'w')

创建csv writer对象

csvwriter = csv.writer(employ_data)
count = 0
for emp in emp_data:
      if count == 0:
             header = emp.keys()
             csvwriter.writerow(header)
             count += 1
      csvwriter.writerow(emp.values())

为了保存内容，请确保关闭文件

employ_data.close()

首先，JSON包含嵌套对象，因此通常不能直接转换为CSV。你需要把它改成这样:

{
    "pk": 22,
    "model": "auth.permission",
    "codename": "add_logentry",
    "content_type": 8,
    "name": "Can add log entry"
},
......]

下面是我的代码来生成CSV:

import csv
import json

x = """[
    {
        "pk": 22,
        "model": "auth.permission",
        "fields": {
            "codename": "add_logentry",
            "name": "Can add log entry",
            "content_type": 8
        }
    },
    {
        "pk": 23,
        "model": "auth.permission",
        "fields": {
            "codename": "change_logentry",
            "name": "Can change log entry",
            "content_type": 8
        }
    },
    {
        "pk": 24,
        "model": "auth.permission",
        "fields": {
            "codename": "delete_logentry",
            "name": "Can delete log entry",
            "content_type": 8
        }
    }
]"""

x = json.loads(x)

f = csv.writer(open("test.csv", "wb+"))

# Write CSV Header, If you dont need that, remove this line
f.writerow(["pk", "model", "codename", "name", "content_type"])

for x in x:
    f.writerow([x["pk"],
                x["model"],
                x["fields"]["codename"],
                x["fields"]["name"],
                x["fields"]["content_type"]])

你会得到如下输出:

pk,model,codename,name,content_type
22,auth.permission,add_logentry,Can add log entry,8
23,auth.permission,change_logentry,Can change log entry,8
24,auth.permission,delete_logentry,Can delete log entry,8

JSON可以表示各种各样的数据结构——JS的“对象”大致类似于Python的dict(带有字符串键)，JS的“数组”大致类似于Python列表，只要最后的“叶子”元素是数字或字符串，你就可以嵌套它们。

CSV本质上只能表示一个2-D表——可选的第一行是“标题”，即“列名”，这可以使表可解释为字典列表，而不是正常的解释，一个列表的列表(同样，“叶子”元素可以是数字或字符串)。

So, in the general case, you can't translate an arbitrary JSON structure to a CSV. In a few special cases you can (array of arrays with no further nesting; arrays of objects which all have exactly the same keys). Which special case, if any, applies to your problem? The details of the solution depend on which special case you do have. Given the astonishing fact that you don't even mention which one applies, I suspect you may not have considered the constraint, neither usable case in fact applies, and your problem is impossible to solve. But please do clarify!

使用pandas库，这就像使用两个命令一样简单!

df = pd.read_json()

read_json将JSON字符串转换为pandas对象(序列或数据帧)。然后:

df.to_csv()

它既可以返回字符串，也可以直接写入csv文件。请参阅to_csv的文档。

根据之前的冗长回答，我们都应该感谢熊猫提供的这条捷径。

关于非结构化JSON，请参阅这个答案。

编辑: 有人问我一个最小的例子:

import pandas as pd

with open('jsonfile.json', encoding='utf-8') as inputfile:
    df = pd.read_json(inputfile)

df.to_csv('csvfile.csv', encoding='utf-8', index=False)

此代码适用于任何给定的json文件

# -*- coding: utf-8 -*-
"""
Created on Mon Jun 17 20:35:35 2019
author: Ram
"""

import json
import csv

with open("file1.json") as file:
    data = json.load(file)



# create the csv writer object
pt_data1 = open('pt_data1.csv', 'w')
csvwriter = csv.writer(pt_data1)

count = 0

for pt in data:

      if count == 0:

             header = pt.keys()

             csvwriter.writerow(header)

             count += 1

      csvwriter.writerow(pt.values())

pt_data1.close()

令人惊讶的是，我发现到目前为止贴在这里的答案都没有正确处理所有可能的场景(例如，嵌套字典，嵌套列表，无值等)。

这个解决方案应该适用于所有场景:

def flatten_json(json):
    def process_value(keys, value, flattened):
        if isinstance(value, dict):
            for key in value.keys():
                process_value(keys + [key], value[key], flattened)
        elif isinstance(value, list):
            for idx, v in enumerate(value):
                process_value(keys + [str(idx)], v, flattened)
        else:
            flattened['__'.join(keys)] = value

    flattened = {}
    for key in json.keys():
        process_value([key], json[key], flattened)
    return flattened