Find keys in nested dictionary

Question

How to find a specific key in a nested dictionary in python without using a for loop and directly accessing that variable.

I have a json file which looks similar to a nested dictionary. How could i find a specific keyword in it?

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'
}
}
}

Now i need to find brand value from this dictionary. Is there any way of retrieving it in single line of code.

Answer 1

I assume, that you have a data structure similar to this:

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

Maybe you could provide an example of you data.

In a nested structure you could use recursion to walk trough. With recursion your code is independent of the data structures depth.

A simple implementation could look like this:

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

I have tested that way:

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

The code should work on arbitrary deep nested dicts. (until you reach the max recursion depth of your python interpreter)

Answer 2

outer_keys  = filter(lambda 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"

are you looking for something like this?