英文:
Getting children and grandchildren in a single query in Gremlin
问题
我目前正在将一些Cypher查询重写为Gremlin。我想创建一个单一查询，针对特定的起始节点，返回以下内容：
- 到子节点 - 基于名为'prob'的边属性。我们希望获取最大概率的最多10个子节点（按边属性'prob'按降序排序）。
- 对于每个子节点，我们想继续并获取最高概率的最多10个孙子节点（类似于第1点）。
我附上了一个显示我们想要的结果的图像 - 但假设为简单起见最多获取2个节点而不是10个。作为结果，我们还希望获取所有子节点和孙子节点的属性。
谢谢！
编辑：
我想出了以下解决方案。也许有人可以指出更好的方法，但似乎该查询返回了正确的结果。
g.V('123')
.inE().order().by('prob', Order.desc).limit(10).outV().as('c')
.project('c', 'gc').by(valueMap(true)).by(inE().order().by('prob', Order.desc).limit(10).outV().valueMap(true).fold())
英文:
I am currently rewriting some of the queries written in Cypher to Gremlin.
I want to create a single query that would return for the specific starting node:
- up to the children - based on the edge property called 'prob.' We want to get up to 10 children with biggest probability (sorted by edge property 'prob' in desc order).
- For every child we would like to continue and get up to 10 grandchildren with the highest probability (similar to point 1).
I attached an image showing the result we would like to get - but with the assumption of getting up to 2 nodes instead of 10 for simplicity. As the result, we would also like to get all the properties of children and grandchildren.
Thank you!
Edit:
I came up with the following solution. Maybe someone point out a better approach but is seems the query returns the correct result.
g.V('123')
.inE().order().by('prob', Order.desc).limit(10).outV().as('c')
.project('c', 'gc').by(valueMap(true)).by(inE().order().by('prob', Order.desc).limit(10).outV().valueMap(true).fold())
答案1
得分: 1
你可以在这种情况下使用 local 步骤，对于每个子节点，您希望有限数量的子子节点。由于我没有您的数据集，以下是一些我认为与您的用例很好匹配的示例。在查看边缘之前，这里是一个基本示例，展示了如何使用 local 可以帮助（我只使用了限制为2，以保持简单）。
g.V('44').
out().limit(2).
local(out().limit(2)).
path().
by('code')
这将产生以下结果：
1 path[SAF, DFW, ASE]
2 path[SAF, DFW, GEG]
3 path[SAF, LAX, YLW]
4 path[SAF, LAX, ASE]
在航空路线数据集中，每条边都有一个 "dist" 属性（路线距离），我们可以使用它来模拟您的用例。
g.V('44').
outE('route').order().by('dist',desc).inV().limit(2).
local(outE('route').order().by('dist',desc).inV().limit(2)).
path().
by('code').
by('dist')
我们可以看到它选择了最长的路线：
1 path[SAF, 708, LAX, 8756, SIN]
2 path[SAF, 708, LAX, 8372, AUH]
3 path[SAF, 549, DFW, 8574, SYD]
4 path[SAF, 549, DFW, 8105, HKG]
英文:
You should be able to use the local step in a case like this where, for each child you want a limited number of grandchildren. As I do not have your data set, here are some examples that I think map well to your use case. Before looking at the edges, here is a basic example that shows how local can help (I used a limit of 2 just to keep things simple).
g.V('44').
out().limit(2).
local(out().limit(2)).
path().
by('code')
This yields
1 path[SAF, DFW, ASE]
2 path[SAF, DFW, GEG]
3 path[SAF, LAX, YLW]
4 path[SAF, LAX, ASE]
In the air routes data set, each edge has a "dist" property (the route distance), we can use that to simulate your use case.
g.V('44').
outE('route').order().by('dist',desc).inV().limit(2).
local(outE('route').order().by('dist',desc).inV().limit(2)).
path().
by('code').
by('dist')
Which we can see picks the longest routes
1 path[SAF, 708, LAX, 8756, SIN]
2 path[SAF, 708, LAX, 8372, AUH]
3 path[SAF, 549, DFW, 8574, SYD]
4 path[SAF, 549, DFW, 8105, HKG]
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