Gremlin遍历从列表中的元素中减去值

使用gods示例的图表并添加以下“amount”属性：

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

下面的遍历基于（Ai-平均值（A））和（双平均值（B））项的计算目标：

遍历的输出：

==>[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:0.6623472535822001]
==>[a1:[v1:v[8192]，v2:v[4096]，a1:[0.32524675]，a2:[0.30577987,0.8416171,0.5247855,0.57484317,0.35161084]，a1m:0.325246713084116，a2m:0.5197273015975952]]
==>[a1:[v1:v[8192]，v2:v[4184]，a1:[0.32524675]，a2:[0.53761715,0.9604127,0.87463444,0.7719325]，a1m:0.3252467513084116，a2m:0.786149188876152]]
==>[a1:[v1:v[8192]，v2:v[4248]，a1:[0.32524675]，a2:[0.7349615,0.80212617,0.6879539]，a1m:0.325246753084116，a2m:0.7416805227597555]]
==>[a1:[v1:v[8192]，v2:v[4264]，a1:[0.32524675]，a2:[0.37226892,0.8902944,0.4158439,0.9709829]，a1m:0.32524675513084116，a2m:0.66234755822001]]
==>[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:0.51972730159752]]
==>[a1:[v1:v[4184]，v2:v[8192]，a1:[0.53761715,0.9604127,0.87463444,0.7719325]，a2:[0.32524675]，a1m:0.786149188876152，a2m:0.325246713084116]]
==>[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:0.51972730159752]]
==>[a1:[v1:v[4248]，v2:v[8192]，a1:[0.7349615,0.80212617,0.6879539]，a2:[0.32524675]，a1m:0.7416805227597555，a2m:0.325246753084116]]
==>[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:0.662347752535822001，a2m:0.51972730159752]]

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有人如何从遍历的这一点计算“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，任何示例帮助都会很好。

既然map中已经有了所有值，让我们从这里开始

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）与展开

a

一样简单，进行数学运算（

ai am

或

（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的值，我相信最终的查询可以简化很多，但最好在另一个线程中处理

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]]