python – SimpleJson处理相同的命名实体

2019年8月4日 213次阅读

我在app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“子类型”被重复,因此负载返回的字典只是“TVActor”而不是列表.反正有没有绕过这个？

最佳答案 定义application / json的
rfc 4627说：

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不应在同一对象内返回多个“subType”名称,并声称它是JSON.

您可以尝试以XML格式请求相同的内容(outputMode = xml)以避免结果中出现歧义或将重复键值转换为列表：

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)

例

text = """{
  "type": "Person",
  "subType": "Athlete",
  "subType": "AwardWinner"
}"""

产量

{u'subType': [u'Athlete', u'AwardWinner'], u'type': u'Person'}
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