Hello,
Thank you. I did upgrade to 2.1.2. When i tried the below statement, the
queries are coming back really fast but since i am using the
allshortestpath, I believe all the nodes including a and c are included in
the count. I just need the nodes that are in the path between a and c.
Is there are way to get ONLY the count of nodes between the nodes a and c?
Thank you for the help.
Raj
On Wednesday, July 2, 2014 1:36:04 PM UTC-7, Michael Hunger wrote:
>
> If you update to 2.1.2 it uses indexes for IN clauses.
>
> And you might be able to do
>
> match (a:newtags) where a.name IN
> ["1225","10091","1921056","11962","1921078","1920809","1921090"]
> match (c:newtags) where c.name IN ["5226","1955"]
>
> match p = allShortestPaths((a)-[*..2]-(c))
> return count(*)
>
>
> Use the profile keyword in the neo4j shell to see if it actually uses an
> index:
>
> like this:
>
> profile
> > match (a:newtags) where a.name IN
> ["1225","10091","1921056","11962","1921078","1920809","1921090"]
> > match (c:newtags) where c.name IN ["5226","1955"]
> > match p = allShortestPaths((a)-[*..2]-(c))
> > return count(*)
> > ;
> +----------+
> | count(*) |
> +----------+
> | 0 |
> +----------+
> 1 row
>
> ColumnFilter
> |
> +EagerAggregation
> |
> +ShortestPath
> |
> +SchemaIndex(0)
> |
> +SchemaIndex(1)
>
>
> +------------------+------+--------+-------------+---------------------------------------------------------------------------------------------------------------------------------------------------------+
> | Operator | Rows | DbHits | Identifiers |
>
> Other |
>
> +------------------+------+--------+-------------+---------------------------------------------------------------------------------------------------------------------------------------------------------+
> | ColumnFilter | 1 | 0 | |
>
> keep columns count(*) |
> | EagerAggregation | 1 | 0 | |
>
> |
> | ShortestPath | 0 | 0 | p |
>
> |
> | SchemaIndex(0) | 0 | 0 | c, c |
>
> Collection(List({ AUTOSTRING7}, { AUTOSTRING8})); :newtags(name) |
> | SchemaIndex(1) | 0 | 7 | a, a | Collection(List({
> AUTOSTRING0}, { AUTOSTRING1}, { AUTOSTRING2}, { AUTOSTRING3}, {
> AUTOSTRING4}, { AUTOSTRING5}, { AUTOSTRING6})); :newtags(name) |
>
> +------------------+------+--------+-------------+---------------------------------------------------------------------------------------------------------------------------------------------------------+
>
> You should also pass in your collections as parameters, so the query can
> be built once and then reused
>
> match (a:newtags) where a.name IN {tags1}
> match (c:newtags) where c.name IN {tags2}
>
> match p = allShortestPaths((a)-[*..2]-(c))
> return count(*)
>
>
>
> Michael
>
> Am 02.07.2014 um 17:10 schrieb jankan99 <[email protected]
> <javascript:>>:
>
> Hello, In the 2.1.1 version, when we do the following cypher it takes a
> long time (more than a minute). When we do the same query using a "With"
> clause, it takes about 6 seconds. We need to get this down to less than a
> second. We think it is slow because all the nodes "a" and "c" are loaded
> in memory before filtering is done using the "where" clause. Is there
> anyway to force Neo to use the indexes on the "name" property to select
> only the nodes in the where clause. We tried USING HINTS but it looks the
> IN clause is not supported for hints.
>
> Cyoher that takes more than a minute
> match (a:newtags)-[:belongs_to]-(b)-[:belongs_to]-(c:newtags)
> where a.name IN
> ["1225","10091","1921056","11962","1921078","1920809","1921090"]
> and c.name IN ["5226","1955"]
> return count(distinct(b));
>
> Cypher using WITH clause takes about 6 seconds
> match (a:newtags) where a.name IN
> ["1225","10091","1921056","11962","1921078","1920809","1921090"]
> with a
> match (a)-[rm1]-(b)-[rm2]-(c:newtags)
> where c.name IN ["5226","1955"]
> return (count(distinct(b))
>
> Some stats:
> The machine has about 2.2 milion nodes. We are using AWS with 7.5 GB RAM.
>
> Any help will be appreciated.
>
> Thank you.
>
> Raj
>
>
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