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
}
}
]
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))
Alec的回答很好，但在存在多层嵌套的情况下行不通。下面是一个支持多层嵌套的修改版本。如果嵌套对象已经指定了自己的键(例如Firebase Analytics / BigTable / BigQuery数据)，它也会使头名称更好一些:
"""Converts JSON with nested fields into a flattened CSV file.
"""
import sys
import json
import csv
import os
import jsonlines
from orderedset import OrderedSet
# from https://stackoverflow.com/a/28246154/473201
def flattenjson( b, prefix='', delim='/', val=None ):
if val is None:
val = {}
if isinstance( b, dict ):
for j in b.keys():
flattenjson(b[j], prefix + delim + j, delim, val)
elif isinstance( b, list ):
get = b
for j in range(len(get)):
key = str(j)
# If the nested data contains its own key, use that as the header instead.
if isinstance( get[j], dict ):
if 'key' in get[j]:
key = get[j]['key']
flattenjson(get[j], prefix + delim + key, delim, val)
else:
val[prefix] = b
return val
def main(argv):
if len(argv) < 2:
raise Error('Please specify a JSON file to parse')
print "Loading and Flattening..."
filename = argv[1]
allRows = []
fieldnames = OrderedSet()
with jsonlines.open(filename) as reader:
for obj in reader:
# print 'orig:\n'
# print obj
flattened = flattenjson(obj)
#print 'keys: %s' % flattened.keys()
# print 'flattened:\n'
# print flattened
fieldnames.update(flattened.keys())
allRows.append(flattened)
print "Exporting to CSV..."
outfilename = filename + '.csv'
count = 0
with open(outfilename, 'w') as file:
csvwriter = csv.DictWriter(file, fieldnames=fieldnames)
csvwriter.writeheader()
for obj in allRows:
# print 'allRows:\n'
# print obj
csvwriter.writerow(obj)
count += 1
print "Wrote %d rows" % count
if __name__ == '__main__':
main(sys.argv)
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))
