Forum Discussion
Exclude duplicate reference from a calculation
Hello the Community,
I have to calculate a percentage of anomaly on the sales of some product.
On the file of anomaly my problem is that I have several type of anomaly on the same reference so on a reference the anomaly is counted several times.
How to make sure that if a reference is present several times, only one is counted in the calculation?
Thanks in advance
Hi Cednok ,
In my understanding, You could create a measure by the following formula:
anomaly percentage = DISTINCTCOUNT([REF])/CALCULATE(COUNT([REF]),FILTER(ALL('Table'),[Article Type]="Sold to"))The final output is shown below:
If it is incorrect, you could try to another measure by the following formula:
anomaly percentage2 = DISTINCTCOUNT([REF])/COUNT([REF])The final output is shown below:
Best Regards,
Community Support Team_ Yalan Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
4 Replies
- amitchandakSuper User
- CednokFrequent Visitor
Thank you but these measurs will tell me how many duplicate references I have, it will not allow me to do the calculation
I may have badly expressed my need
I have to calculate a percentage of anomaly on the reference of articles but I have several anomalies on the same references.
But if several anomalies on the same reference = 1 anomaly
For example on the file I should have 8 real anomalyFor examples At the moment I have 10 articles "sold to", on the file anomaly you can see 5 anomalies but on 3 references the calculation should be 3/10 and not 5/10
How to make so that the calculation is taken into account only on the distinct references ?
- v-yalanwu-msftCommunity Support
Hi Cednok ,
In my understanding, You could create a measure by the following formula:
anomaly percentage = DISTINCTCOUNT([REF])/CALCULATE(COUNT([REF]),FILTER(ALL('Table'),[Article Type]="Sold to"))The final output is shown below:
If it is incorrect, you could try to another measure by the following formula:
anomaly percentage2 = DISTINCTCOUNT([REF])/COUNT([REF])The final output is shown below:
Best Regards,
Community Support Team_ Yalan Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.- CednokFrequent Visitor
Thank you very much