Skip to main content
Announcements
Join us at Qlik Connect for 3 magical days of learning, networking,and inspiration! REGISTER TODAY and save!
cancel
Showing results for 
Search instead for 
Did you mean: 
Not applicable

Finding the minium of an series of values

Hi Qlik,

I have a list of sites that each do a series of deliveries to mulitple destinations.

I have a table set up with Source [Source Name] as the dimension showing the average cost avg[Total Cost] of delivery to each destination [Destination Name].

Looking at opportunity cost I want to establish as a varible what the cheapest site to ship to is. However merely entering min(Total Cost) gives me back the cheapest delievry overall rather than the average of the site with lowest cost.

Is there a way to create an array within the equation of average total cost to each site and then select the minimum average value?

Regards,

Chris

1 Solution

Accepted Solutions
Gysbert_Wassenaar

Perhaps like this: min(aggr(avg([Total Cost]),[Destination Name], [Source Name]))


talk is cheap, supply exceeds demand

View solution in original post

6 Replies
Gysbert_Wassenaar

Perhaps like this: min(aggr(avg([Total Cost]),[Destination Name], [Source Name]))


talk is cheap, supply exceeds demand
Not applicable
Author

Thank you Gysbert, worked like a charm.

swuehl
MVP
MVP

If your request is resolved, then please close this thread:

Qlik Community Tip: Marking Replies as Correct or Helpful

Not applicable
Author

As a follow up, I'm now trying to use this to find the next best alternative. So far I have:

if(count(distinct([Destination Name]))>1,

min({$<[Total Cost]={"<>'[Best route]'"}>} [Best Route]))

Where best route is the previous equation.

swuehl
MVP
MVP

In general, try to keep a thread to a single request.

Since it's quite similar, maybe

if(count(distinct([Destination Name]))>1,

min(aggr(avg([Total Cost]),[Destination Name], [Source Name]), 2)

)

Not applicable
Author

Apologies Swuehl, will start a new one next time.

Thank you your answer worked perfectly.