Question

我在app引擎中使用Alchemy API，因此我使用simplejson库来解析响应。问题是响应的条目具有名称

 {
    "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，因此加载返回的字典只是“TVActor”而不是列表。无论如何都要绕过这个？

Answer 1

定义application/json的{{3}}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不应在同一对象中返回多个"subType"名称，并声称它是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)

实施例

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

输出

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

Answer 2

The rfc 4627 for application/json media type建议使用唯一键，但不会明确禁止它们：

对象中的名称应唯一。

来自rfc 2119：

应该这个词，或形容词“ RECOMMENDED”，表示在那里
  在特定情况下可能存在正当理由而忽略了
  特定项目，但必须理解全部含义并
  在选择其他路线之前，请仔细权衡。

这是一个已知的问题。

您可以通过修改重复键或将其保存到数组中来解决此问题。您可以根据需要使用此代码。

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)

SimpleJson处理相同的命名实体

2 个答案:

实施例

输出