抽出キー

Question

次のようになります。私は抽出する必要があり

{ 
"took": 34, 
"timed_out": false, 
"_shards": { 
"total": 1760, 
"successful": 1760, 
"failed": 0 
}, 
"hits": { 
"total": 551, 
"max_score": 0.30685282, 
"hits": [ 
{ 
"_index": "instances-i-0034438e-2016.10.10-08:23:51", 
"_type": "type_name", 
"_id": "1", 
"_score": 0.30685282, 
"_source": { 
"ansible.isv_alias": "ruiba", 
"ansible": { 
"isv_alias": "ruiba", 
"WEB_URL": "sin01-cloud.trial.sentinelcloud.com/ruiba" 
} 
} 
} 
, 
{ 
"_index": "instances-i-0034438e-2016.10.11-08:23:54", 
"_type": "type_name", 
"_id": "1", 
"_score": 0.30685282, 
"_source": { 
"ansible.isv_alias": "aike3", 
"ansible": { 
"isv_alias": "aike3", 
"WEB_URL": "sin01-cloud.trial.cloud.com/aike3" 
} 
} 
} 
, 
{ 
"_index": "instances-i-883sf38e-2016.10.12-08:23:45", 
"_type": "type_name", 
"_id": "1", 
"_score": 0.30685282, 
"_source": { 
"ansible.isv_alias": "beijing", 
"ansible": { 
"isv_alias": "beijing", 
"WEB_URL": "sin01-cloud.trial.cloud.com/beijing" 
} 
} 
} 
. 
. 
. 
. 
so on 

返されたdictのWEB_URLをLISTに格納します。私たちが実際に移入さLISTでWEB_URLの値の重複エントリがないことを確認することができた場合

はまた、私は好奇心{これが第二の部分ですけれども}

Answer 1

必要にWEB_URLの値を格納しますdictionary-

data = response_from_es['hits'] 
list_of_urls = [] 
for item in data['hits']: 
    list_of_urls.append(item['_source']['ansible']['WEB_URL']) 

print list_of_urls 

を反復処理するとlist_of_urls unique作るために - >set(list_of_urls)

あなたは、同様の答えを受け入れるべき