Question

如何在python中的嵌套字典中查找特定键而不使用for循环并直接访问该变量。

我有一个类似于嵌套字典的json文件。我怎么能在其中找到特定的关键字？

data={
      u'session': {
      u'new': True,
      u'sessionId': u'amzn1.echo-api.session.fae41185-1c00-4463-b890-0833c32a2b43',
      u'user': {
      u'userId': u'amzn1.ask.account.AHEEJXVACGJ2XZV4QRNFSIW7LDHAXEUB7WTZXC6777CLHLS4ITEWD6RMQLNXG5ECXFP3WRAQDH53EZUZ2GOFC52SNZ5NI5L7LXG3HMS6ORA2DNVT3TFYL5N5WE4GTSFM4PS7TGYXXQYCMOCRVBEPXBNHV2G6DPIUP3XPHCSIKJF26SOSZZAD4TRLHJA66JPEVDWJMRBH7AL2WUA'
},
      u'application': {
      u'applicationId': u'amzn1.ask.skill.ec977129-bcaf-462c-85ce-03c1f7a2b1f8'
 }
},
u'version': u'1.0',
u'request': {
u'locale': u'en-IN',
u'timestamp': u'2018-01-12T10:11:46Z',
u'dialogState': u'STARTED',
u'intent': {
  u'slots': {
    u'product': {
      u'name': u'product',
      u'confirmationStatus': u'NONE'
    },
    u'subcategories': {
      u'resolutions': {
        u'resolutionsPerAuthority': [
          {
            u'status': {
              u'code': u'ER_SUCCESS_MATCH'
            },
            u'values': [
              {
                u'value': {
                  u'name': u'milk',
                  u'id': u'ecbdb882ae865a07d87611437fda0772'
                }
              }
            ],
            u'authority': u'amzn1.er-authority.echo-sdk.amzn1.ask.skill.ec977129-bcaf-462c-85ce-03c1f7a2b1f8_Certification.subcategories'
          }
        ]
      },
      u'name': u'subcategories',
      u'value': u'milk',
      u'confirmationStatus': u'NONE'
    },
    u'brand': {
      u'resolutions': {
        u'resolutionsPerAuthority': [
          {
            u'status': {
              u'code': u'ER_SUCCESS_MATCH'
            },
            u'values': [
              {
                u'value': {
                  u'name': u'nestle',
                  u'id': u'099e3726deeeb0f04f86d49093eebbb2'
                }
              }
            ],
            u'authority': u'amzn1.er-authority.echo-sdk.amzn1.ask.skill.ec977129-bcaf-462c-85ce-03c1f7a2b1f8_Certification.brand'
          }
        ]
      },
      u'name': u'brand',
      u'value': u'nestle',
      u'confirmationStatus': u'NONE'
    },
    u'categories': {
      u'name': u'categories',
      u'confirmationStatus': u'NONE'
    },
    u'more_n_repeat': {
      u'name': u'more_n_repeat',
      u'confirmationStatus': u'NONE'
    }
  },
  u'name': u'user_says',
  u'confirmationStatus': u'NONE'
},
u'requestId': u'amzn1.echo-api.request.432d4866-f781-418a-88a8-5d2b8d341409',
u'type': u'IntentRequest'
},
u'context': {
u'AudioPlayer': {
  u'playerActivity': u'IDLE'
},
u'System': {
  u'device': {
    u'deviceId': u'amzn1.ask.device.AHJQTUWHP2J7MS7WFE45RNWOO32KO4DPTY4HKNU66YPLG3BOFXRFVA32JEILOKUADFOGC4JCTM4CFOPN6MJP2BEQNF3TPOKCPGLHYEDJBLYS5QOIBPKD7SZALTCVKK2YUP7WWUA5XHC2ATFS4MYLXA7I4UVQ',
    u'supportedInterfaces': {
      u'AudioPlayer': {

      }
    }
  },
  u'application': {
    u'applicationId': u'amzn1.ask.skill.ec977129-bcaf-462c-85ce-03c1f7a2b1f8'
  },
  u'apiAccessToken': u'eyJ0eXAiOiJKV1QiLCJhbGciOiJSUzI1NiIsImtpZCI6IjEifQ.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.DsQSWRNaAlNyzdOyCxaYa3n193RxYJCul0RFc-prCDXxpzIYRxJ6wrZkAS7BCSezK-6hcL8wuiFGdVtQPQk57mL8TBjxwczyc_6vHMjeQEzbcIXGE-xLx-lifNxIpH6w9YQNLavMsLIEEsMP0X1dMXDiBXBAe9A6neb0EdTRi6mdJaUC6Lvx6YnZrY1Xu8scm3vqi47BHKYsGaO508m9vZgVIjmz3st5LHW3Uc5svsK0T3P3aew0iH7K9pzpfm1mRF6POd4jLpyoikdLY4HEDQOC9RSLN4W0SvTw1-y8UEerkQolHaJWQdv86F1dqKjlbaF1zRAmWRVYPPVp7gqFag',
  u'user': {
    u'userId': u'amzn1.ask.account.AHEEJXVACGJ2XZV4QRNFSIW7LDHAXEUB7WTZXC6777CLHLS4ITEWD6RMQLNXG5ECXFP3WRAQDH53EZUZ2GOFC52SNZ5NI5L7LXG3HMS6ORA2DNVT3TFYL5N5WE4GTSFM4PS7TGYXXQYCMOCRVBEPXBNHV2G6DPIUP3XPHCSIKJF26SOSZZAD4TRLHJA66JPEVDWJMRBH7AL2WUA'
  },
  u'apiEndpoint': u'https://api.eu.amazonalexa.com'
}
}
}

现在我需要从这本词典中找到品牌价值。有没有办法在单行代码中检索它。

Answer 1

我假设你有一个类似的数据结构：

data = {
    'a': 42,
    'b': {
        'ba': 23,
        'bb': {
            'bba': 420
        }
    }
}

也许您可以提供一个数据示例。

在嵌套结构中，您可以使用递归来走槽。使用递归，您的代码与数据结构深度无关。

一个简单的实现可能如下所示：

def recursive_lookup(k, d):
    if k in d:
        return d[k]
    for v in d.values():
        if isinstance(v, dict):
            return recursive_lookup(k, v)
    return None

我测试过这种方式：

print('a', recursive_lookup('a', data))
print('ba', recursive_lookup('ba', data))
print('bba', recursive_lookup('bba', data))

代码应该适用于任意深层嵌套的dicts。（直到达到python解释器的最大递归深度）

Answer 2

outer_keys  = filter(lamba x: my_key in x, my_dict.keys())

result = my_dict[next(outer_keys)][my_key]

Answer 3

varName = "orange"
if varName in dict_data:
    print "i found " + varName + ":" + dict_data[varName]
else
    print varName + " not found"

你正在寻找这样的东西吗？

在嵌套字典中查找键

3 个答案: