我有一个JSON文件,我想转换为CSV文件。我如何用Python做到这一点?

我试着:

import json
import csv

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

f = open('data.csv')
csv_file = csv.writer(f)
for item in data:
    csv_file.writerow(item)

f.close()

然而,这并没有起作用。我正在使用Django和我收到的错误是:

`file' object has no attribute 'writerow'`

然后我尝试了以下方法:

import json
import csv

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

f = open('data.csv')
csv_file = csv.writer(f)
for item in data:
    f.writerow(item)  # ← changed

f.close()

然后得到错误:

`sequence expected`

样本json文件:

[{
        "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
        }
    }, {
        "pk": 4,
        "model": "auth.permission",
        "fields": {
            "codename": "add_group",
            "name": "Can add group",
            "content_type": 2
        }
    }, {
        "pk": 10,
        "model": "auth.permission",
        "fields": {
            "codename": "add_message",
            "name": "Can add message",
            "content_type": 4
        }
    }
]

当前回答

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

其他回答

使用pandas中的json_normalize:

在名为test.json的文件中使用来自OP的示例数据。 这里使用了Encoding ='utf-8',但在其他情况下可能不需要。 下面的代码利用了pathlib库。 .open是pathlib的一个方法。 也适用于非windows路径。 使用pandas.to_csv(…)将数据保存为csv文件。

import pandas as pd
# As of Pandas 1.01, json_normalize as pandas.io.json.json_normalize is deprecated and is now exposed in the top-level namespace.
# from pandas.io.json import json_normalize
from pathlib import Path
import json

# set path to file
p = Path(r'c:\some_path_to_file\test.json')

# read json
with p.open('r', encoding='utf-8') as f:
    data = json.loads(f.read())

# create dataframe
df = pd.json_normalize(data)

# dataframe view
 pk            model  fields.codename           fields.name  fields.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
  4  auth.permission        add_group         Can add group                    2
 10  auth.permission      add_message       Can add message                    4

# save to csv
df.to_csv('test.csv', index=False, encoding='utf-8')

CSV输出:

pk,model,fields.codename,fields.name,fields.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
4,auth.permission,add_group,Can add group,2
10,auth.permission,add_message,Can add message,4

嵌套更重的JSON对象的资源:

所以答案: 用python平化JSON数组 如何平嵌套的JSON递归,与平坦JSON 如何json_normalize一个列与nan 使用pandas将一列字典拆分为单独的列 有关其他相关问题,请参阅json_normalize标记。

这段代码应该适用于您,假设您的JSON数据在一个名为data. JSON的文件中。

import json
import csv

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

with open("data.csv", "w") as file:
    csv_file = csv.writer(file)
    for item in data:
        fields = list(item['fields'].values())
        csv_file.writerow([item['pk'], item['model']] + fields)

不幸的是,我没有足够的声誉来为@Alec McGail的惊人回答做出小小的贡献。 我正在使用Python3,我需要将映射转换为@Alexis R注释后面的列表。

另外,我发现csv作者添加了一个额外的CR文件(我有一个空行每一行与数据在csv文件)。根据@Jason R. Coombs对这个帖子的回答,解决方法非常简单: CSV在Python中添加了一个额外的回车

您只需将lineterminator='\n'参数添加到csv.writer。它将是:csv_w = csv。Writer (out_file, lineterminator='\n')

我已经尝试了很多建议的解决方案(也熊猫没有正确地规范化我的JSON),但真正好的是正确解析JSON数据来自Max Berman。

我写了一个改进,以避免每一行都有新列 在解析期间将其放置到现有列。 如果只有一个数据存在,则将值存储为字符串,如果该列有更多值,则将值存储为列表。

它有一个输入。Json文件作为输入,并输出一个output.csv。

import json
import pandas as pd

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, v, flattened)
                # process_value(keys + [str(idx)], v, flattened)
        else:
            key1 = '__'.join(keys)
            if not flattened.get(key1) is None:
                if isinstance(flattened[key1], list):
                    flattened[key1] = flattened[key1] + [value]
                else:
                    flattened[key1] = [flattened[key1]] + [value]
            else:
                flattened[key1] = value

    flattened = {}
    for key in json.keys():
        k = key
        # print("Key: " + k)
        process_value([key], json[key], flattened)
    return flattened

try:
    f = open("input.json", "r")
except:
    pass
y = json.loads(f.read())
flat = flatten_json(y)
text = json.dumps(flat)
df = pd.read_json(text)
df.to_csv('output.csv', index=False, encoding='utf-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!