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gremlin - Gremlin遍历从列表中的元素中减去值

问题描述

使用诸神的图表并添加以下“金额”属性：

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

下面的遍历基于https://neo4j.com/docs/graph-algorithms/current/algorithms/similarity-pearson/，目的是计算 (Ai - mean(A)) 和 (Bi - mean(B) ) 条款：

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)))

遍历输出：

==>[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]]

有人将如何从遍历中的这一点计算“a1 - a1m”和“a2 - a2m”？这里的问题是从列表中的每个元素中减去一个值，然后返回差异列表，任何对示例的帮助都会很棒。

标签： gremlinjanusgraph

解决方案

由于您已经在 map 中拥有所有值，让我们从那里开始。

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]

从每个单个值 (ai) 中减去平均值 (am) 就像展开一样简单a，做数学运算 ( ai-amor (am-ai)*(-1)) 并将它们折叠在一起：

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，对于这两个值，它只是另一个投影：

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]]

我想还有几个步骤可以得出最终值，我相信最终查询可以简化很多，但最好在另一个线程中处理。

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