Yes, it works! Thanks! > On Dec 9, 2015, at 10:41 AM, Yingyi Bu <[email protected]> wrote: > > This probably is the query you want: > > for $ts in $hashtags > for $t in $ts > group by $tag:=$t with $t > return {"g": $tag, count($t)} > > > On Wed, Dec 9, 2015 at 10:34 AM, Jianfeng Jia <[email protected]> > wrote: > >> Hi devs, >> >> Here is my use case, each tweets has a set of hashtags. Here is the >> hashtags example for the first five tweets: >> [ {{ "Samibeigi" }}, {{ "LeadwithGiants" }}, {{ "MountainView", >> "Healthcare", "Job" }}, {{ "SanFrancisco", "job", "NettempsJobs", "IT", >> "Hiring", "CareerArc" }}, {{ "BeGreat" }} ] >> I want to calculate the most frequent hashtag in all tweets. >> >> I could generate a internal group count as following >> >> let $inner := for $x in $hashtags >> return >> for $xx in $x >> group by $xx with $xx >> return { "g": $xx[0], "c": count($xx)} >> >> It will return a list of list, >> [ { "g": "Samibeigi", "c": 1 } ] >> [ { "g": "LeadwithGiants", "c": 1 } ] >> [ { "g": "Healthcare", "c": 1 }, { "g": "Job", "c": 1 }, { "g": >> "MountainView", "c": 1 } ] >> [ { "g": "CareerArc", "c": 1 }, { "g": "Hiring", "c": 1 }, { "g": "IT", >> "c": 1 }, { "g": "NettempsJobs", "c": 1 }, { "g": "SanFrancisco", "c": 1 }, >> { "g": "job", "c": 1 } ] >> [ { "g": "BeGreat", "c": 1 } ] >> >> I would expect to add a flatten function to the list which will give me a >> list of record, then I can groupby the “g". Do we have such AQL functions? >> Thank you. >> >> Best, >> >> Jianfeng Jia >> PhD Candidate of Computer Science >> University of California, Irvine >> >>
Best, Jianfeng Jia PhD Candidate of Computer Science University of California, Irvine
