Cypher：分层排序

Question

我是Cypher的新用户。我能够创建一个地理网络（从世界到大陆，从国家到地区）及其人口。

你可以用这个命令重现或检查链接到控制台： http://console.neo4j.org/r/dkh90c

CREATE (n1:Geo {name:'World'}), 
(n2:Geo {name:'EMEA'})-[:BELONG_TO]->(n1), 
(n4:Geo {name:'NORAM'})-[:BELONG_TO]->(n1), 
(n5:Geo {name:'Middle East'})-[:BELONG_TO]->(n2), 
(n6:Geo {name:'Africa'})-[:BELONG_TO]->(n2), 
(n7:Geo {name:'Europe'})-[:BELONG_TO]->(n2), 
(n8:Geo {name:'France'})-[:BELONG_TO]->(n7), 
(n9:Geo {name:'Germany'})-[:BELONG_TO]->(n7), 
(n10:Geo {name:'Italy'})-[:BELONG_TO]->(n7), 
(n11:Geo {name:'United Kingdom'})-[:BELONG_TO]->(n7), 
(n12:Geo {name:'England'})-[:BELONG_TO]->(n11), 
(n13:Geo {name:'Scotland'})-[:BELONG_TO]->(n11), 
(n14:Geo {name:'Wales'})-[:BELONG_TO]->(n11), 
(n15:Geo {name:'Northern Ireland'})-[:BELONG_TO]->(n11), 
(n16:Geo {name:'United Arab Emirates'})-[:BELONG_TO]->(n5), 
(n17:Geo {name:'South Africa'})-[:BELONG_TO]->(n6), 
(n18:Geo {name:'Canada'})-[:BELONG_TO]->(n4), 
(n19:Geo {name:'United States of America'})-[:BELONG_TO]->(n4), 
(n20:Geo {name:'Mexico'})-[:BELONG_TO]->(n4), 

(:Population {year:'2014',amount:66.1})-[:LIVE_IN]->(n8), 
(:Population {year:'2014',amount:81.2})-[:LIVE_IN]->(n9), 
(:Population {year:'2013',amount:59.83})-[:LIVE_IN]->(n10), 
(:Population {year:'2011',amount:53.01})-[:LIVE_IN]->(n12), 
(:Population {year:'2011',amount:5.295})-[:LIVE_IN]->(n13), 
(:Population {year:'2011',amount:3.063})-[:LIVE_IN]->(n14), 
(:Population {year:'2011',amount:1.811})-[:LIVE_IN]->(n15), 
(:Population {year:'2013',amount:9.346})-[:LIVE_IN]->(n16), 
(:Population {year:'2013',amount:52.98})-[:LIVE_IN]->(n17), 
(:Population {year:'2013',amount:35.16})-[:LIVE_IN]->(n18), 
(:Population {year:'2014',amount:318.9})-[:LIVE_IN]->(n19), 
(:Population {year:'2013',amount:122.3})-[:LIVE_IN]->(n20) 

我也能够计算的总人口与此命令每个地区：

MATCH (n:Population)-[r:LIVE_IN]->(g1:Geo)-[:BELONG_TO*0..]->(g2:Geo) 
RETURN g2.name AS Geography, SUM(toFloat(n.amount)) AS Population 
ORDER BY Population DESC 

但是，我不满意结果的排序：

如您所见，美国插入欧洲，中东和非洲地区。在显示结果之前，是否有办法按“BELONG_TO”层次排序？另外，尽管我的所有数据都只有3位小数，但我不明白为什么SUM（）命令返回疯狂的小数位数。

感谢您的帮助。

Answer 1

当然，你应该能够通过路径的长度进行排序：

MATCH path=(n:Population)-[r:LIVE_IN]->(g1:Geo)-[:BELONG_TO*0..]->(g2:Geo) 
RETURN g2.name AS Geography, SUM(toFloat(n.amount)) AS Population 
ORDER BY length(path) ASC, Population DESC 

我觉得你得到很多，因为floating point math小数。我想你应该能够做到：

ROUND(SUM(toFloat(n.amount)) * 1000.0)/1000.0 

编辑：

你说得对，你需要添加的路径长度由它来订购：

MATCH path=(n:Population)-[r:LIVE_IN]->(g1:Geo)-[:BELONG_TO*1..]->(g2:Geo) 
RETURN g1.name AS Geography, SUM(toFloat(n.amount)) AS Population, length(path) AS path_length 
ORDER BY length(path) ASC, Population DESC 

是的，你肯定是正确的，你的道路需要以g2为世界结束。你可以通过匹配(g2:Geo {name: 'World'})来实现这一点，类似于我在评论中提出的建议，或者如果你将来会有更多的根节点（也许我们将殖民月球或火星！），你可以这样做：

MATCH path=(n:Population)-[r:LIVE_IN]->(g1:Geo)-[:BELONG_TO*]->(g2:Geo) 
WHERE NOT((g2:Geo)-[:BELONG_TO]->()) 
RETURN g1.name AS Geography, SUM(toFloat(n.amount)) AS Population, length(path) AS path_length 
ORDER BY length(path) ASC, Population DESC 

这意味着我们只想要g2不属于任何东西的路径。否则，g2可能是，例如，Europe