我在应用程序引擎中使用Alchemy API,所以我使用simplejson库来解析响应。问题是响应的条目具有sme名称
{
"status": "OK",
"usage": "By accessing AlchemyAPI or using information generated by AlchemyAPI, you are agreeing to be bound by the AlchemyAPI Terms of Use: http://www.alchemyapi.com/company/terms.html",
"url": "",
"language": "english",
"entities": [
{
"type": "Person",
"relevance": "0.33",
"count": "1",
"text": "Michael Jordan",
"disambiguated": {
"name": "Michael Jordan",
"subType": "Athlete",
"subType": "AwardWinner",
"subType": "BasketballPlayer",
"subType": "HallOfFameInductee",
"subType": "OlympicAthlete",
"subType": "SportsLeagueAwardWinner",
"subType": "FilmActor",
"subType": "TVActor",
"dbpedia": "http://dbpedia.org/resource/Michael_Jordan",
"freebase": "http://rdf.freebase.com/ns/guid.9202a8c04000641f8000000000029161",
"umbel": "http://umbel.org/umbel/ne/wikipedia/Michael_Jordan",
"opencyc": "http://sw.opencyc.org/concept/Mx4rvViVq5wpEbGdrcN5Y29ycA",
"yago": "http://mpii.de/yago/resource/Michael_Jordan"
}
}
]
}所以问题是"subType“是重复的,所以加载返回的只是"TVActor”而不是列表。有什么办法可以绕过这个问题吗?
发布于 2011-10-20 05:51:32
定义application/json的rfc 4627表示:
An object is an unordered collection of zero or more name/value pairs和:
The names within an object SHOULD be unique.这意味着AlchemyAPI不应该在同一个对象中返回多个"subType"名称,并声明它是一个JSON。
您可以尝试以XML格式(outputMode=xml)请求相同的内容,以避免结果中的歧义,或者将重复的键值转换为列表:
import simplejson as json
from collections import defaultdict
def multidict(ordered_pairs):
"""Convert duplicate keys values to lists."""
# read all values into lists
d = defaultdict(list)
for k, v in ordered_pairs:
d[k].append(v)
# unpack lists that have only 1 item
for k, v in d.items():
if len(v) == 1:
d[k] = v[0]
return dict(d)
print json.JSONDecoder(object_pairs_hook=multidict).decode(text)示例
text = """{
"type": "Person",
"subType": "Athlete",
"subType": "AwardWinner"
}"""输出
{u'subType': [u'Athlete', u'AwardWinner'], u'type': u'Person'}发布于 2019-01-15 22:46:07
The rfc 4627 for application/json media type建议使用唯一键,但并未明确禁止:
对象中的名称应该是唯一的。
来自rfc 2119
应该推荐这个词,或者形容词“”,意思是
在特定情况下可能存在有效的理由来忽略
特定项目,但必须了解全部含义,并
在选择不同的课程之前仔细权衡过。
这是一个已知的问题。
您可以通过修改重复密钥来解决这个问题,也可以将其保存到数组中。如果您愿意,您可以使用此代码。
import json
def parse_object_pairs(pairs):
"""
This function get list of tuple's
and check if have duplicate keys.
if have then return the pairs list itself.
but if haven't return dict that contain pairs.
>>> parse_object_pairs([("color": "red"), ("size": 3)])
{"color": "red", "size": 3}
>>> parse_object_pairs([("color": "red"), ("size": 3), ("color": "blue")])
[("color": "red"), ("size": 3), ("color": "blue")]
:param pairs: list of tuples.
:return dict or list that contain pairs.
"""
dict_without_duplicate = dict()
for k, v in pairs:
if k in dict_without_duplicate:
return pairs
else:
dict_without_duplicate[k] = v
return dict_without_duplicate
decoder = json.JSONDecoder(object_pairs_hook=parse_object_pairs)
str_json_can_be_with_duplicate_keys = '{"color": "red", "size": 3, "color": "red"}'
data_after_decode = decoder.decode(str_json_can_be_with_duplicate_keys)https://stackoverflow.com/questions/7825261
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