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!

Alec的回答很好，但在存在多层嵌套的情况下行不通。下面是一个支持多层嵌套的修改版本。如果嵌套对象已经指定了自己的键(例如Firebase Analytics / BigTable / BigQuery数据)，它也会使头名称更好一些:

"""Converts JSON with nested fields into a flattened CSV file.
"""

import sys
import json
import csv
import os

import jsonlines

from orderedset import OrderedSet

# from https://stackoverflow.com/a/28246154/473201
def flattenjson( b, prefix='', delim='/', val=None ):
  if val is None:
    val = {}

  if isinstance( b, dict ):
    for j in b.keys():
      flattenjson(b[j], prefix + delim + j, delim, val)
  elif isinstance( b, list ):
    get = b
    for j in range(len(get)):
      key = str(j)

      # If the nested data contains its own key, use that as the header instead.
      if isinstance( get[j], dict ):
        if 'key' in get[j]:
          key = get[j]['key']

      flattenjson(get[j], prefix + delim + key, delim, val)
  else:
    val[prefix] = b

  return val

def main(argv):
  if len(argv) < 2:
    raise Error('Please specify a JSON file to parse')

  print "Loading and Flattening..."
  filename = argv[1]
  allRows = []
  fieldnames = OrderedSet()
  with jsonlines.open(filename) as reader:
    for obj in reader:
      # print 'orig:\n'
      # print obj
      flattened = flattenjson(obj)
      #print 'keys: %s' % flattened.keys()
      # print 'flattened:\n'
      # print flattened
      fieldnames.update(flattened.keys())
      allRows.append(flattened)

  print "Exporting to CSV..."
  outfilename = filename + '.csv'
  count = 0
  with open(outfilename, 'w') as file:
    csvwriter = csv.DictWriter(file, fieldnames=fieldnames)
    csvwriter.writeheader()
    for obj in allRows:
      # print 'allRows:\n'
      # print obj
      csvwriter.writerow(obj)
      count += 1

  print "Wrote %d rows" % count



if __name__ == '__main__':
  main(sys.argv)

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()

一个通用的解决方案，将任何json列表的平面对象转换为csv。

传递输入。Json文件作为命令行的第一个参数。

import csv, json, sys

input = open(sys.argv[1])
data = json.load(input)
input.close()

output = csv.writer(sys.stdout)

output.writerow(data[0].keys())  # header row

for row in data:
    output.writerow(row.values())

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 } }
]

我假设您的JSON文件将解码为字典列表。首先，我们需要一个将JSON对象扁平化的函数:

def flattenjson(b, delim):
    val = {}
    for i in b.keys():
        if isinstance(b[i], dict):
            get = flattenjson(b[i], delim)
            for j in get.keys():
                val[i + delim + j] = get[j]
        else:
            val[i] = b[i]
            
    return val

在JSON对象上运行这段代码的结果:

flattenjson({
    "pk": 22, 
    "model": "auth.permission", 
    "fields": {
      "codename": "add_message", 
      "name": "Can add message", 
      "content_type": 8
    }
  }, "__")

is

{
    "pk": 22, 
    "model": "auth.permission", 
    "fields__codename": "add_message", 
    "fields__name": "Can add message", 
    "fields__content_type": 8
}

对JSON对象输入数组中的每个dict应用此函数后:

input = map(lambda x: flattenjson( x, "__" ), input)

并查找相关的列名:

columns = [x for row in input for x in row.keys()]
columns = list(set(columns))

在CSV模块中运行这个并不难:

with open(fname, 'wb') as out_file:
    csv_w = csv.writer(out_file)
    csv_w.writerow(columns)

    for i_r in input:
        csv_w.writerow(map(lambda x: i_r.get(x, ""), columns))