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Hello Qlik Community,
I have seen various ways to group data using aggr and I can't seem to get this to work.
In a chart I have a dimension (UserID). I have a count of (Submissions). I can see how many Submissions each UserID has submitted. I want to group by these totals. For example I want to see how many Users submitted 1, or submitted 2, or submitted 47, etc...
The ideal result would be seeing this in a histogram or other chart where I could see the Total Amount of Users per Submission totals.
Example:
1,000 users had 1 submissions, 500 users has 2 submissions, etc.
Any ideas? I am open to Data load editor, or set analysis - whichever would be best practice in your mind.
Here is an example of what I was trying in set analysis, but couldn't get it to show up right in a table or chart. Please note I had some filters for the Submission type.
=Aggr(Count({$<[SOURCE]={'sourcevalue'},[TYPE]={'typevalue'}>} DISTINCT SUBMISSIONID),USERID)
-Thanks!
Hi,
I would suggest you to create a table which will consist of only two columns. UserID and number of submissions. Something like this.
SubmissionsCount:
Load
UserID
count(SUBMISSIONID) as numbeOfSubmissions
resident <your primary table>
group by UserId;
and than just simply use histogram chart where you will use field numberOfSubmission.
That's it . Hope this will help you.
Are you using QlikView or Qlik Sense ?
Hi,
in dimension :
aggr(count distinct submissionID,userID)
in expression :
count(distinct userId)
and you order by the expression
but may be it's not axactly what you want ?
regards
Hi,
I would suggest you to create a table which will consist of only two columns. UserID and number of submissions. Something like this.
SubmissionsCount:
Load
UserID
count(SUBMISSIONID) as numbeOfSubmissions
resident <your primary table>
group by UserId;
and than just simply use histogram chart where you will use field numberOfSubmission.
That's it . Hope this will help you.
Thanks David,
This was a very clean solution using the Data Load Editor. Much easier then trying the same thing in Set Analysis. Your response is much appreciated.
Best regards,
-Derek