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Not applicable

Calculating last value by date/status

Hey folks,

In the following example, I want to sum up the number of issues that ended up at status for a cyclic group containing date and month.

Script:

Data:

LOAD * Inline

[

  Date, Sequence, Issue, Status

  2016-06-01, 1, 100, Open

  2016-06-01, 2, 100, In Progress

2016-06-02, 1, 100, Open

  2016-07-02, 1, 100, Closed

  2016-06-01, 1, 200, Open

  2016-07-02, 1, 200, In Progress

  2016-07-02, 2, 200, Closed

];

Left Join (Data)

LOAD

  *,

  Month(Date) AS Month

Resident Data;

The resulting chart will have one of the two sets of values, depending on what the cyclic group is set to:

Cyclic Group = DateDate

Status

Sum
2016-06-01In Progress1
2016-06-01Open1
2016-06-02Open1
2016-07-02Closed2

Or:

Cyclic Group = MonthStatusSum
JunOpen2
JulClosed2

What's the best way to go about doing this?


Thanks,

Ken

Message was edited by: Ken van Mulder Missed adding line #7 of the script to include data for 2016-06-02.

4 Replies
Kushal_Chawda

like this

Data: 

LOAD *, monthname(Date) as Monthyear Inline 

  Date, Sequence, Issue, Status 

  2016-06-01, 1, 100, Open 

  2016-06-01, 2, 100, In Progress 

  2016-07-02, 1, 100, Closed 

  2016-06-01, 1, 200, Open 

  2016-07-02, 1, 200, In Progress 

  2016-07-02, 2, 200, Closed 

]; 

Now create the Cyclic group , add Date & MonthYear .

Create straight table

Dimension:

CyclicGroup

Status

Expression:

Count(Status)

MindaugasBacius
Partner - Specialist III
Partner - Specialist III

Please find the qvw attachment.

The results looks like this:

Screenshot_1.jpg

tamilarasu
Champion
Champion

Hi Ken,

Have a look at the attached file.

Not applicable
Author

Hello,

Thanks for the answers - much appreciated, but unfortunately they're missing the core of the problem.

I only want to count the issue once for the status that it ends the selected time period. See the example tables for the expected output.

I've worked out a way to get it working using a calculated value for relevant record per time period, but I'm hoping there is a better solution.

I've attached a file showing the sample solution to date.

Thanks,

Ken