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
}
}
]
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。
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))
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')
