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!

修改了Alec McGail的答案，以支持包含列表的JSON

    def flattenjson(self, mp, delim="|"):
            ret = []
            if isinstance(mp, dict):
                    for k in mp.keys():
                            csvs = self.flattenjson(mp[k], delim)
                            for csv in csvs:
                                    ret.append(k + delim + csv)
            elif isinstance(mp, list):
                    for k in mp:
                            csvs = self.flattenjson(k, delim)
                            for csv in csvs:
                                    ret.append(csv)
            else:
                    ret.append(mp)

            return ret

谢谢!

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))

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)

如果我们考虑下面的例子，将json格式的文件转换为csv格式的文件。

{
 "item_data" : [
      {
        "item": "10023456",
        "class": "100",
        "subclass": "123"
      }
      ]
}

下面的代码将转换json文件(data3. xml)。Json)转换为CSV文件(data3.csv)。

import json
import csv
with open("/Users/Desktop/json/data3.json") as file:
    data = json.load(file)
    file.close()
    print(data)

fname = "/Users/Desktop/json/data3.csv"

with open(fname, "w", newline='') as file:
    csv_file = csv.writer(file)
    csv_file.writerow(['dept',
                       'class',
                       'subclass'])
    for item in data["item_data"]:
         csv_file.writerow([item.get('item_data').get('dept'),
                            item.get('item_data').get('class'),
                            item.get('item_data').get('subclass')])

上面提到的代码已经在本地安装的pycharm中执行，它已经成功地将json文件转换为csv文件。希望这有助于转换文件。

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!