解析嵌套JSON并将其写入CSV
解析嵌套JSON并将其写入CSV
提问于 2013-12-06 12:55:33
我在努力解决这个问题。我有一个JSON文件，需要将它输出到CSV，如果结构是平面的，没有深嵌套项，那么它很好。
但在这种情况下，嵌套的RACES使我很不舒服。
我将如何以这样的格式获取数据：
VENUE, COUNTRY, ITW, RACES__NO, RACES__TIME对于对象中的每个对象和每个种族？
{
"1": {
"VENUE": "JOEBURG",
"COUNTRY": "HAE",
"ITW": "XAD",
"RACES": {
"1": {
"NO": 1,
"TIME": "12:35"
},
"2": {
"NO": 2,
"TIME": "13:10"
},
"3": {
"NO": 3,
"TIME": "13:40"
},
"4": {
"NO": 4,
"TIME": "14:10"
},
"5": {
"NO": 5,
"TIME": "14:55"
},
"6": {
"NO": 6,
"TIME": "15:30"
},
"7": {
"NO": 7,
"TIME": "16:05"
},
"8": {
"NO": 8,
"TIME": "16:40"
}
}
},
"2": {
"VENUE": "FOOBURG",
"COUNTRY": "ABA",
"ITW": "XAD",
"RACES": {
"1": {
"NO": 1,
"TIME": "12:35"
},
"2": {
"NO": 2,
"TIME": "13:10"
},
"3": {
"NO": 3,
"TIME": "13:40"
},
"4": {
"NO": 4,
"TIME": "14:10"
},
"5": {
"NO": 5,
"TIME": "14:55"
},
"6": {
"NO": 6,
"TIME": "15:30"
},
"7": {
"NO": 7,
"TIME": "16:05"
},
"8": {
"NO": 8,
"TIME": "16:40"
}
}
}, ...
}我想这样输出给CSV：
VENUE, COUNTRY, ITW, RACES__NO, RACES__TIME
JOEBERG, HAE, XAD, 1, 12:35
JOEBERG, HAE, XAD, 2, 13:10
JOEBERG, HAE, XAD, 3, 13:40
...
...
FOOBURG, ABA, XAD, 1, 12:35
FOOBURG, ABA, XAD, 2, 13:10所以，首先我得到正确的钥匙：
self.keys = self.data.keys()
keys = ["DATA_KEY"]
for key in self.keys:
if type(self.data[key]) == dict:
for k in self.data[key].keys():
if k not in keys:
if type(self.data[key][k]) == unicode:
keys.append(k)
elif type(self.data[key][k]) == dict:
self.subkey = k
for sk in self.data[key][k].values():
for subkey in sk.keys():
subkey = "%s__%s" % (self.subkey, subkey)
if subkey not in keys:
keys.append(subkey)然后添加数据：
但是怎么做呢？
这应该是一个有趣的，对你熟练的执法人员。;-)
回答 1
Stack Overflow用户
回答已采纳
发布于 2013-12-06 15:43:12
我只收集第一个对象的键，然后假设其余的格式是一致的。
下面的代码还将嵌套对象限制为一个；当有多个对象时，您没有指定应该发生什么。拥有两个或多个相同长度的嵌套结构可以工作(将这些嵌套结构“压缩”在一起)，但是如果您有不同长度的结构，则需要明确选择如何处理这些结构；如何使用空列进行压缩，或者写出这些条目的乘积( an行，每次找到B项时重复来自A的信息)。
import csv
from operator import itemgetter
with open(outputfile, 'wb') as outf:
writer = None # will be set to a csv.DictWriter later
for key, item in sorted(data.items(), key=itemgetter(0)):
row = {}
nested_name, nested_items = '', {}
for k, v in item.items():
if not isinstance(v, dict):
row[k] = v
else:
assert not nested_items, 'Only one nested structure is supported'
nested_name, nested_items = k, v
if writer is None:
# build fields for each first key of each nested item first
fields = sorted(row)
# sorted keys of first item in key sorted order
nested_keys = sorted(sorted(nested_items.items(), key=itemgetter(0))[0][1])
fields.extend('__'.join((nested_name, k)) for k in nested_keys)
writer = csv.DictWriter(outf, fields)
writer.writeheader()
for nkey, nitem in sorted(nested_items.items(), key=itemgetter(0)):
row.update(('__'.join((nested_name, k)), v) for k, v in nitem.items())
writer.writerow(row)对于您的示例输入，这将生成：
COUNTRY,ITW,VENUE,RACES__NO,RACES__TIME
HAE,XAD,JOEBURG,1,12:35
HAE,XAD,JOEBURG,2,13:10
HAE,XAD,JOEBURG,3,13:40
HAE,XAD,JOEBURG,4,14:10
HAE,XAD,JOEBURG,5,14:55
HAE,XAD,JOEBURG,6,15:30
HAE,XAD,JOEBURG,7,16:05
HAE,XAD,JOEBURG,8,16:40
ABA,XAD,FOOBURG,1,12:35
ABA,XAD,FOOBURG,2,13:10
ABA,XAD,FOOBURG,3,13:40
ABA,XAD,FOOBURG,4,14:10
ABA,XAD,FOOBURG,5,14:55
ABA,XAD,FOOBURG,6,15:30
ABA,XAD,FOOBURG,7,16:05
ABA,XAD,FOOBURG,8,16:40页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/20424473
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