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!

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()

这工作得相对较好。 它将json压缩成csv文件。 嵌套元素被管理:)

这是python 3的

import json

o = json.loads('your json string') # Be careful, o must be a list, each of its objects will make a line of the csv.

def flatten(o, k='/'):
    global l, c_line
    if isinstance(o, dict):
        for key, value in o.items():
            flatten(value, k + '/' + key)
    elif isinstance(o, list):
        for ov in o:
            flatten(ov, '')
    elif isinstance(o, str):
        o = o.replace('\r',' ').replace('\n',' ').replace(';', ',')
        if not k in l:
            l[k]={}
        l[k][c_line]=o

def render_csv(l):
    ftime = True

    for i in range(100): #len(l[list(l.keys())[0]])
        for k in l:
            if ftime :
                print('%s;' % k, end='')
                continue
            v = l[k]
            try:
                print('%s;' % v[i], end='')
            except:
                print(';', end='')
        print()
        ftime = False
        i = 0

def json_to_csv(object_list):
    global l, c_line
    l = {}
    c_line = 0
    for ov in object_list : # Assumes json is a list of objects
        flatten(ov)
        c_line += 1
    render_csv(l)

json_to_csv(o)

享受。

我对丹提出的解决方案感到困惑，但这对我来说很管用:

import json
import csv 

f = open('test.json')
data = json.load(f)
f.close()

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

for item in data:
  f.writerow([item['pk'], item['model']] + item['fields'].values())

“测试的地方。Json”包含以下内容:

[ 
{"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 } }
]

您可以使用此代码将json文件转换为csv文件 读取文件后，我将对象转换为熊猫数据框架，然后将其保存为CSV文件

import os
import pandas as pd
import json
import numpy as np

data = []
os.chdir('D:\\Your_directory\\folder')
with open('file_name.json', encoding="utf8") as data_file:    
     for line in data_file:
        data.append(json.loads(line))

dataframe = pd.DataFrame(data)        
## Saving the dataframe to a csv file
dataframe.to_csv("filename.csv", encoding='utf-8',index= False)

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!