neo4j cypher

Neo4j的最常見用途之一是構建實時推薦引擎，一個共同的主題是它們利用大量不同的數據來提出有趣的推薦。

例如， 在此視頻中， 阿曼達（Amanda）展示了約會網站如何通過社交聯系開始，然后介紹熱情，位置和其他一些東西，從而構建實時推薦引擎。

Graph Aware有一個簡潔的框架 ，可以幫助您使用Java構建自己的推薦引擎，我很好奇Cypher版本的外觀。

這是示例圖：

CREATE(m:Person:Male {name:'Michal', age:30}),(d:Person:Female {name:'Daniela', age:20}),(v:Person:Male {name:'Vince', age:40}),(a:Person:Male {name:'Adam', age:30}),(l:Person:Female {name:'Luanne', age:25}),(c:Person:Male {name:'Christophe', age:60}),(lon:City {name:'London'}),(mum:City {name:'Mumbai'}),(m)-[:FRIEND_OF]->(d),(m)-[:FRIEND_OF]->(l),(m)-[:FRIEND_OF]->(a),(m)-[:FRIEND_OF]->(v),(d)-[:FRIEND_OF]->(v),(c)-[:FRIEND_OF]->(v),(d)-[:LIVES_IN]->(lon),(v)-[:LIVES_IN]->(lon),(m)-[:LIVES_IN]->(lon),(l)-[:LIVES_IN]->(mum);

我們想向“亞當”推薦一些潛在的朋友，因此我們查詢的第一層是找到他的朋友，因為其中肯定有一些潛在的朋友：

MATCH (me:Person {name: "Adam"}) MATCH (me)-[:FRIEND_OF]-()-[:FRIEND_OF]-(potentialFriend) RETURN me, potentialFriend, COUNT(*) AS friendsInCommon==> +--------------------------------------------------------------------------------------+ ==> | me | potentialFriend | friendsInCommon | ==> +--------------------------------------------------------------------------------------+ ==> | Node[1007]{name:"Adam",age:30} | Node[1006]{name:"Vince",age:40} | 1 | ==> | Node[1007]{name:"Adam",age:30} | Node[1005]{name:"Daniela",age:20} | 1 | ==> | Node[1007]{name:"Adam",age:30} | Node[1008]{name:"Luanne",age:25} | 1 | ==> +--------------------------------------------------------------------------------------+ ==> 3 rows

該查詢為我們提供了潛在朋友的列表以及我們有多少個共同的朋友。

現在我們有了一些潛在的朋友，讓我們開始為他們每個人建立一個排名。 一個可以吸引潛在朋友的指標是，如果他們和我們生活在同一地點，那么可以將其添加到查詢中：

MATCH (me:Person {name: "Adam"}) MATCH (me)-[:FRIEND_OF]-()-[:FRIEND_OF]-(potentialFriend)WITH me, potentialFriend, COUNT(*) AS friendsInCommonRETURN me,potentialFriend,SIZE((potentialFriend)-[:LIVES_IN]->()<-[:LIVES_IN]-(me)) AS sameLocation==> +-----------------------------------------------------------------------------------+ ==> | me | potentialFriend | sameLocation | ==> +-----------------------------------------------------------------------------------+ ==> | Node[1007]{name:"Adam",age:30} | Node[1006]{name:"Vince",age:40} | 0 | ==> | Node[1007]{name:"Adam",age:30} | Node[1005]{name:"Daniela",age:20} | 0 | ==> | Node[1007]{name:"Adam",age:30} | Node[1008]{name:"Luanne",age:25} | 0 | ==> +-----------------------------------------------------------------------------------+ ==> 3 rows

接下來，我們將通過比較每個節點的標簽來檢查Adams的潛在朋友是否與他具有相同的性別。 我們提供了“性別”和“性別”標簽。

MATCH (me:Person {name: "Adam"}) MATCH (me)-[:FRIEND_OF]-()-[:FRIEND_OF]-(potentialFriend)WITH me, potentialFriend, COUNT(*) AS friendsInCommonRETURN me,potentialFriend,SIZE((potentialFriend)-[:LIVES_IN]->()<-[:LIVES_IN]-(me)) AS sameLocation,LABELS(me) = LABELS(potentialFriend) AS gender==> +--------------------------------------------------------------------------------------------+ ==> | me | potentialFriend | sameLocation | gender | ==> +--------------------------------------------------------------------------------------------+ ==> | Node[1007]{name:"Adam",age:30} | Node[1006]{name:"Vince",age:40} | 0 | true | ==> | Node[1007]{name:"Adam",age:30} | Node[1005]{name:"Daniela",age:20} | 0 | false | ==> | Node[1007]{name:"Adam",age:30} | Node[1008]{name:"Luanne",age:25} | 0 | false | ==> +--------------------------------------------------------------------------------------------+ ==> 3 rows

接下來，讓我們計算亞當和他的潛在朋友之間的年齡差異：

MATCH (me:Person {name: "Adam"}) MATCH (me)-[:FRIEND_OF]-()-[:FRIEND_OF]-(potentialFriend)WITH me, potentialFriend, COUNT(*) AS friendsInCommonRETURN me,potentialFriend,SIZE((potentialFriend)-[:LIVES_IN]->()<-[:LIVES_IN]-(me)) AS sameLocation,abs( me.age - potentialFriend.age) AS ageDifference,LABELS(me) = LABELS(potentialFriend) AS gender,friendsInCommon==> +--------------------------------------------------------------------------------------+ ==> | me | potentialFriend | sameLocation | ageDifference | gender | friendsInCommon | ==> +--------------------------------------------------------------------------------------+ ==> | Node[1007]{name:"Adam",age:30} | Node[1006]{name:"Vince",age:40} | 0 | 10.0 | true | 1 | ==> | Node[1007]{name:"Adam",age:30} | Node[1005]{name:"Daniela",age:20} | 0 | 10.0 | false | 1 | ==> | Node[1007]{name:"Adam",age:30} | Node[1008]{name:"Luanne",age:25} | 0 | 5.0 | false | 1 | ==> +--------------------------------------------------------------------------------------+ ==> 3 rows

現在，讓我們進行一些過濾，以擺脫與亞當已經成為朋友的人–推薦這些人沒有多大意義！

MATCH (me:Person {name: "Adam"}) MATCH (me)-[:FRIEND_OF]-()-[:FRIEND_OF]-(potentialFriend)WITH me, potentialFriend, COUNT(*) AS friendsInCommonWITH me,potentialFriend,SIZE((potentialFriend)-[:LIVES_IN]->()<-[:LIVES_IN]-(me)) AS sameLocation,abs( me.age - potentialFriend.age) AS ageDifference,LABELS(me) = LABELS(potentialFriend) AS gender,friendsInCommonWHERE NOT (me)-[:FRIEND_OF]-(potentialFriend)RETURN me,potentialFriend,SIZE((potentialFriend)-[:LIVES_IN]->()<-[:LIVES_IN]-(me)) AS sameLocation,abs( me.age - potentialFriend.age) AS ageDifference,LABELS(me) = LABELS(potentialFriend) AS gender,friendsInCommon==> +---------------------------------------------------------------------------------------+ ==> | me | potentialFriend | sameLocation | ageDifference | gender | friendsInCommon | ==> +---------------------------------------------------------------------------------------+ ==> | Node[1007]{name:"Adam",age:30} | Node[1006]{name:"Vince",age:40} | 0 | 10.0 | true | 1 | ==> | Node[1007]{name:"Adam",age:30} | Node[1005]{name:"Daniela",age:20} | 0 | 10.0 | false | 1 | ==> | Node[1007]{name:"Adam",age:30} | Node[1008]{name:"Luanne",age:25} | 0 | 5.0 | false | 1 | ==> +---------------------------------------------------------------------------------------+ ==> 3 rows

在這種情況下，我們實際上并未將任何人過濾掉，但是對于其他一些人，我們會看到潛在朋友數量的減少。

我們的最后一步是為每個我們認為對提出朋友建議很重要的功能評分。

如果人們居住在與亞當相同的地方或性別相同，我們將給滿分10分，否則給0分。 對于ageDifference和friendsInCommon，我們將應用對數曲線，以使這些值不會對我們的最終分數產生不成比例的影響。 我們將使用ParetoScoreTransfomer中定義的公式來執行此操作：

public <OUT> float transform(OUT item, float score) {if (score < minimumThreshold) {return 0;}double alpha = Math.log((double) 5) / eightyPercentLevel;double exp = Math.exp(-alpha * score);return new Double(maxScore * (1 - exp)).floatValue();}

現在，對于我們完整的推薦查詢：

MATCH (me:Person {name: "Adam"}) MATCH (me)-[:FRIEND_OF]-()-[:FRIEND_OF]-(potentialFriend)WITH me, potentialFriend, COUNT(*) AS friendsInCommonWITH me,potentialFriend,SIZE((potentialFriend)-[:LIVES_IN]->()<-[:LIVES_IN]-(me)) AS sameLocation,abs( me.age - potentialFriend.age) AS ageDifference,LABELS(me) = LABELS(potentialFriend) AS gender,friendsInCommonWHERE NOT (me)-[:FRIEND_OF]-(potentialFriend)WITH potentialFriend,// 100 -> maxScore, 10 -> eightyPercentLevel, friendsInCommon -> score (from the formula above)100 * (1 - exp((-1.0 * (log(5.0) / 10)) * friendsInCommon)) AS friendsInCommon,sameLocation * 10 AS sameLocation,-1 * (10 * (1 - exp((-1.0 * (log(5.0) / 20)) * ageDifference))) AS ageDifference,CASE WHEN gender THEN 10 ELSE 0 END as sameGenderRETURN potentialFriend,{friendsInCommon: friendsInCommon,sameLocation: sameLocation,ageDifference:ageDifference,sameGender: sameGender} AS parts,friendsInCommon + sameLocation + ageDifference + sameGender AS score ORDER BY score DESC==> +---------------------------------------------------------------------------------------+ ==> | potentialFriend | parts | score | ==> +---------------------------------------------------------------------------------------+ ==> | Node[1006]{name:"Vince",age:40} | {friendsInCommon -> 14.86600774792154, sameLocation -> 0, ageDifference -> -5.52786404500042, sameGender -> 10} | 19.33814370292112 | ==> | Node[1008]{name:"Luanne",age:25} | {friendsInCommon -> 14.86600774792154, sameLocation -> 0, ageDifference -> -3.312596950235779, sameGender -> 0} | 11.55341079768576 | ==> | Node[1005]{name:"Daniela",age:20} | {friendsInCommon -> 14.86600774792154, sameLocation -> 0, ageDifference -> -5.52786404500042, sameGender -> 0} | 9.33814370292112 | ==> +----------------------------------------------------------------------------------------+

最終查詢還不錯-唯一真正復雜的部分是對數曲線計算。 用戶定義的功能將在將來發揮作用。

這種方法的好處是我們不必走出密碼的道路，因此，如果您對Java不滿意，仍然可以進行實時建議！ 另一方面，推薦引擎的不同部分混合在一起，因此要查看整個管道并不像使用圖形感知框架那樣容易。

下一步是將其應用于Twitter圖形，并在此提供關注者建議。

翻譯自: https://www.javacodegeeks.com/2015/03/neo4j-generating-real-time-recommendations-with-cypher.html

neo4j cypher

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