Gremlin遍歷從列表中的元素中減去值

Question

使用諸神示例的圖表並添加以下'amount'屬性：

rand = new Random()
g.withSack {rand.nextFloat()}.E().property('amount',sack())

下面的遍歷基於https://neo4j.com/docs/graph-algorithms/current/algorithms/similarity-pearson/ ，目標是計算（Ai - mean（A））和（Bi - mean（B） ）條款：

g.V().match(
    __.as('v1').outE().valueMap().select('amount').fold().as('e1'),
    __.as('v1').V().as('v2'),
    __.as('v2').outE().valueMap().select('amount').fold().as('e2'),
    __.as('v1').outE().inV().dedup().fold().as('v1n'),
    __.as('v2').outE().inV().dedup().fold().as('v2n')
    ).
    where('v1',neq('v2').and(without('v1n'))).
    where('v2',without('v1n')).
    project('v1','v2','a1','a2','a1m','a2m').
        by(select('v1')).
        by(select('v2')).
        by(select('e1')).
        by(select('e2')).
        by(select('e1').unfold().mean()).
        by(select('e2').unfold().mean()).
    where(select('a1').unfold().count().is(gt(0))).
    where(select('a2').unfold().count().is(gt(0)))

遍歷輸出：

==>[a1:[v1:v[4096],v2:v[4248],a1:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],a2:[0.7349615,0.80212617,0.6879539],a1m:0.5197273015975952,a2m:0.7416805227597555]] ==>[a1:[v1:v[4096],v2:v[4264],a1:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],a2:[0.37226892,0.8902944,0.4158439,0.9709829],a1m:0.5197273015975952,a2m:0.6623475253582001]] ==>[a1:[v1:v[8192],v2:v[4096],a1:[0.32524675],a2:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],a1m:0.32524675130844116,a2m:0.5197273015975952]] ==>[a1:[v1:v[8192],v2:v[4184],a1:[0.32524675],a2:[0.53761715,0.9604127,0.87463444,0.7719325],a1m:0.32524675130844116,a2m:0.786149188876152]] ==>[a1:[v1:v[8192],v2:v[4248],a1:[0.32524675],a2:[0.7349615,0.80212617,0.6879539],a1m:0.32524675130844116,a2m:0.7416805227597555]] ==>[a1:[v1:v[8192],v2:v[4264],a1:[0.32524675],a2:[0.37226892,0.8902944,0.4158439,0.9709829],a1m:0.32524675130844116,a2m:0.6623475253582001]] ==>[a1:[v1:v[4184],v2:v[4096],a1:[0.53761715,0.9604127,0.87463444,0.7719325],a2:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],a1m:0.786149188876152,a2m:0.5197273015975952]] ==>[a1:[v1:v[4184],v2:v[8192],a1:[0.53761715,0.9604127,0.87463444,0.7719325],a2:[0.32524675],a1m:0.786149188876152,a2m:0.32524675130844116]] ==>[a1:[v1:v[4248],v2:v[4096],a1:[0.7349615,0.80212617,0.6879539],a2:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],a1m:0.7416805227597555,a2m:0.5197273015975952]] ==>[a1:[v1:v[4248],v2:v[8192],a1:[0.7349615,0.80212617,0.6879539],a2:[0.32524675],a1m:0.7416805227597555,a2m:0.32524675130844116]] ==>[a1:[v1:v[4264],v2:v[4096],a1:[0.37226892,0.8902944,0.4158439,0.9709829],a2:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],a1m:0.6623475253582001,a2m:0.5197273015975952]]

怎么會在遍歷中從這一點計算'a1-a1m'和'a2-a2m'？ 這里的問題是從列表中的每個元素中減去一個值，然后返回差異列表，對示例的任何幫助都會很棒。

Answer 1

既然你已經擁有了地圖中的所有值，那就讓我們從那里開始吧。

gremlin> __.inject(['a1': [0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],
......1>            'a2': [0.7349615,0.80212617,0.6879539],
......2>            'a1m': 0.5197273015975952,
......3>            'a2m': 0.7416805227597555])
==>[a1:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],a2:[0.7349615,0.80212617,0.6879539],a1m:0.5197273015975952,a2m:0.7416805227597555]

從每個單個值（ai）中減去平均值（am）就像展開a一樣簡單，進行數學運算（ ai-am或(am-ai)*(-1) ）並將它們折疊在一起：

gremlin> __.inject(['a1': [0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],
......1>            'a2': [0.7349615,0.80212617,0.6879539],
......2>            'a1m': 0.5197273015975952,
......3>            'a2m': 0.7416805227597555]).
......4>    sack(assign).
......5>      by(select('a1m')).
......6>    select('a1').unfold().
......7>    sack(minus).
......8>    sack(mult).
......9>      by(constant(-1)).
.....10>    sack().fold()
==>[-0.2139474315975952,0.3218897984024048,0.0050581984024048,0.0551158684024048,-0.1681164615975952]

因此，對於這兩個值，它只是另一個投影：

gremlin> __.inject(['a1': [0.30577987,0.8416171,0.5247855,0.57484317,0.35161084],
......1>            'a2': [0.7349615,0.80212617,0.6879539],
......2>            'a1m': 0.5197273015975952,
......3>            'a2m': 0.7416805227597555]).
......4>    project('a','b').
......5>      by(sack(assign).
......6>           by(select('a1m')).
......7>         select('a1').unfold().
......8>         sack(minus).
......9>         sack(mult).
.....10>           by(constant(-1)).
.....11>         sack().fold()).
.....12>      by(sack(assign).
.....13>           by(select('a2m')).
.....14>         select('a2').unfold().
.....15>         sack(minus).
.....16>         sack(mult).
.....17>           by(constant(-1)).
.....18>         sack().fold())
==>[a:[-0.2139474315975952,0.3218897984024048,0.0050581984024048,0.0551158684024048,-0.1681164615975952],b:[-0.0067190227597555,0.0604456472402445,-0.0537266227597555]]

我想會有更多的步驟來提出最終值，我確信最終的查詢可以簡化很多，但最好在另一個線程中處理。