Question

我有一个等级（比如1到103）和一个密码中每个等级的节点数。

day.name    rank    count
12/14/2016  1   1
12/14/2016  2   2
12/14/2016  3   3
12/14/2016  4   10
...
12/14/2016  11  20000
12/14/2016  12  20500
...
12/14/2016  21  15000
12/14/2016  22  15000

我希望在列中显示摘要数据，例如排名为1到3，排名为4到10的计数等。

在此示例中，所需的结果为

'rank_1_to_3'           'rank_4_to_10'      'rank_11_to_20' 'rank_21_to_100'
---------------------------------------------------------------------------
6                       10                   40500             30000

这里是Cypher：

match (d:Domain {name:'hotelguides.com'}) 
match (c:Temp {name:"top30k_rank_1_to_100"}) 
match (day:Day {name:'2016-12-22'}) 
match (day)-[]-(kua:KeywordURLAssociation)-[]-(d) 
match (k:Keyword)-[]-(kua)-[]-(r:Rank) 
with day, kua, r, k  return day.name, r.name, count(k)

Answer 1

我认为此查询可以帮助您入门。我将把它应用到你的特定领域。实质上，您可以使用filter将结果集合过滤到具有这些结果子集的另一个集合中。对于每个较小的结果，您可以使用reduce来汇总该集合。像这样......

with [
  {name:'12/14/2016', rank:1, count:1},
  {name:'12/14/2016', rank:2, count:2},
  {name:'12/14/2016', rank:3, count:3},
  {name:'12/14/2016', rank:4, count:10},
  {name:'12/14/2016', rank:11, count:20000},
  {name:'12/14/2016', rank:12, count:20500},
  {name:'12/14/2016', rank:21, count:15000},
  {name:'12/14/2016', rank:22, count:15000}
] as days
return reduce(sum_c=0, d in filter(d in days where d.rank < 4) | sum_c + d.count) as rank_1_3
, reduce(sum_c=0, d in filter(d in days where d.rank > 3 and d.rank < 11) | sum_c + d.count) as rank_4_to_10
, reduce(sum_c=0, d in filter(d in days where d.rank > 10 and d.rank < 21) | sum_c + d.count) as rank_11_to_20
, reduce(sum_c=0, d in filter(d in days where d.rank > 20 and d.rank < 101) | sum_c + d.count) as rank_21_100

如何压平密码结果以在列中显示摘要数据？

1 个答案: