正如在前面的回答中提到的，将json转换为csv的困难在于json文件可以包含嵌套字典，因此是多维数据结构，而csv是2D数据结构。但是，将多维结构转换为csv的一个好方法是使用多个主键连接在一起的csv。

在你的例子中，第一个csv输出的列是“pk”，“model”，“fields”。“pk”和“model”的值很容易获得，但因为“fields”列包含一个字典，它应该是它自己的csv，因为“codename”似乎是主键，你可以使用作为“fields”的输入来完成第一个csv。第二个csv包含来自“fields”列的字典，以codename作为主键，可用于将两个csv绑定在一起。

这是一个解决方案，为您的json文件转换嵌套字典2 csv。

import csv
import json

def readAndWrite(inputFileName, primaryKey=""):
    input = open(inputFileName+".json")
    data = json.load(input)
    input.close()

    header = set()

    if primaryKey != "":
        outputFileName = inputFileName+"-"+primaryKey
        if inputFileName == "data":
            for i in data:
                for j in i["fields"].keys():
                    if j not in header:
                        header.add(j)
    else:
        outputFileName = inputFileName
        for i in data:
            for j in i.keys():
                if j not in header:
                    header.add(j)

    with open(outputFileName+".csv", 'wb') as output_file:
        fieldnames = list(header)
        writer = csv.DictWriter(output_file, fieldnames, delimiter=',', quotechar='"')
        writer.writeheader()
        for x in data:
            row_value = {}
            if primaryKey == "":
                for y in x.keys():
                    yValue = x.get(y)
                    if type(yValue) == int or type(yValue) == bool or type(yValue) == float or type(yValue) == list:
                        row_value[y] = str(yValue).encode('utf8')
                    elif type(yValue) != dict:
                        row_value[y] = yValue.encode('utf8')
                    else:
                        if inputFileName == "data":
                            row_value[y] = yValue["codename"].encode('utf8')
                            readAndWrite(inputFileName, primaryKey="codename")
                writer.writerow(row_value)
            elif primaryKey == "codename":
                for y in x["fields"].keys():
                    yValue = x["fields"].get(y)
                    if type(yValue) == int or type(yValue) == bool or type(yValue) == float or type(yValue) == list:
                        row_value[y] = str(yValue).encode('utf8')
                    elif type(yValue) != dict:
                        row_value[y] = yValue.encode('utf8')
                writer.writerow(row_value)

readAndWrite("data")

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

解决这个问题的简单方法是:

创建一个新的Python文件，如:json_to_csv.py

添加以下代码:

import csv, json, sys
#if you are not using utf-8 files, remove the next line
sys.setdefaultencoding("UTF-8")
#check if you pass the input file and output file
if sys.argv[1] is not None and sys.argv[2] is not None:

    fileInput = sys.argv[1]
    fileOutput = sys.argv[2]

    inputFile = open(fileInput)
    outputFile = open(fileOutput, 'w')
    data = json.load(inputFile)
    inputFile.close()

    output = csv.writer(outputFile)

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

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

添加代码后，保存文件并在终端上运行:

Python json_to_csv.py input.txt output.csv

我希望这对你有帮助。

韩国歌手组合!

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

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')

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!