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 } }
]

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)

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

这是@MikeRepass回答的修改。此版本将CSV写入文件，适用于Python 2和Python 3。

import csv,json
input_file="data.json"
output_file="data.csv"
with open(input_file) as f:
    content=json.load(f)
try:
    context=open(output_file,'w',newline='') # Python 3
except TypeError:
    context=open(output_file,'wb') # Python 2
with context as file:
    writer=csv.writer(file)
    writer.writerow(content[0].keys()) # header row
    for row in content:
        writer.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))