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

这工作得相对较好。 它将json压缩成csv文件。 嵌套元素被管理:)

这是python 3的

import json

o = json.loads('your json string') # Be careful, o must be a list, each of its objects will make a line of the csv.

def flatten(o, k='/'):
    global l, c_line
    if isinstance(o, dict):
        for key, value in o.items():
            flatten(value, k + '/' + key)
    elif isinstance(o, list):
        for ov in o:
            flatten(ov, '')
    elif isinstance(o, str):
        o = o.replace('\r',' ').replace('\n',' ').replace(';', ',')
        if not k in l:
            l[k]={}
        l[k][c_line]=o

def render_csv(l):
    ftime = True

    for i in range(100): #len(l[list(l.keys())[0]])
        for k in l:
            if ftime :
                print('%s;' % k, end='')
                continue
            v = l[k]
            try:
                print('%s;' % v[i], end='')
            except:
                print(';', end='')
        print()
        ftime = False
        i = 0

def json_to_csv(object_list):
    global l, c_line
    l = {}
    c_line = 0
    for ov in object_list : # Assumes json is a list of objects
        flatten(ov)
        c_line += 1
    render_csv(l)

json_to_csv(o)

享受。

其他回答

由于数据看起来是字典格式,因此似乎应该实际使用csv.DictWriter()来实际输出带有适当标题信息的行。这将使转换更容易处理。然后fieldnames参数将正确地设置顺序,而第一行的输出作为标题将允许稍后由csv.DictReader()读取和处理。

例如,Mike Repass使用

output = csv.writer(sys.stdout)

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

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

不过,只需将初始设置更改为 输出= csv。DictWriter数据(文件集,字段名= [0]. keys ())

注意,由于字典中元素的顺序没有定义,您可能必须显式地创建字段名条目。一旦你这样做了,writerow就可以工作了。然后写操作就像最初显示的那样工作。

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))

这不是一个很聪明的方法,但我也遇到过同样的问题,这对我来说很有效:

import csv

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

new_data = []

for i in data:
   flat = {}
   names = i.keys()
   for n in names:
      try:
         if len(i[n].keys()) > 0:
            for ii in i[n].keys():
               flat[n+"_"+ii] = i[n][ii]
      except:
         flat[n] = i[n]
   new_data.append(flat)  

f = open(filename, "r")
writer = csv.DictWriter(f, new_data[0].keys())
writer.writeheader()
for row in new_data:
   writer.writerow(row)
f.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())

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)