4

我在应用引擎中使用 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”被重复了，所以负载返回的字典只是“TVActor”而不是一个列表。有没有办法解决这个问题？

4

2 回答 2

6

定义的rfc 4627application/json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"subType"在同一对象内返回多个名称并声称它是 JSON。

您可以尝试以 XML 格式 ( outputMode=xml) 请求相同的内容，以避免结果中的歧义或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)

例子

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

输出

{u'subType': [u'Athlete', u'AwardWinner'], u'type': u'Person'}
于 2011-10-19T21:51:31.510 回答
1

application/json用于媒体类型的 rfc 4627建议使用唯一键，但并未明确禁止它们：

对象中的名称应该是唯一的。

来自RFC 2119：

应该 这个词，或形容词“推荐”，意味着
在特定情况下可能存在忽略特定项目的正当理由
，但在选择不同的课程之前，必须理解并
仔细权衡全部含义。

这是一个已知的问题。

您可以通过修改重复键来解决这个问题，或者将他保存到数组中。如果需要，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)
于 2019-01-15T14:46:06.703 回答