Supplies are limited. Contact info@espc.tech right away to save your spot before the conference sells out.
Get your discountScore big with last-minute savings on the final tickets to FabCon Vienna. Secure your discount
Hi all,
Need your help. I'm trying to flag customers that made multiple purchases in the same day. The purchase ID must be within 3 numbers of each other. So the same customer on the same day makes purchase ID 100, 103, and 104 should all be flagged as consecutive. Below is a sample table and the consecitive column is what I'm trying to calculate.
Customer | Date | Purchase ID | Consecutive |
A | 4/30/2025 | 1125 | No |
B | 5/20/2025 | 1244 | No |
C | 3/5/2025 | 1003 | Yes |
C | 3/5/2025 | 1004 | Yes |
C | 3/5/2025 | 1006 | Yes |
D | 6/10/2025 | 1356 | No |
D | 6/11/2025 | 1357 | No |
E | 6/12/2025 | 1391 | No |
E | 6/12/2025 | 1395 | No |
Solved! Go to Solution.
Hi @buttercream
Consider building a new Column in your table with below:
Hi,
This calculated column formula works
=if(CALCULATE(COUNTROWS(Data),FILTER(Data,Data[Customer]=EARLIER(Data[Customer])))=1,"No",if(or(Data[Purchase ID]-CALCULATE(MAX(Data[Purchase ID]),FILTER(Data,Data[Customer]=EARLIER(Data[Customer])&&Data[Date]=EARLIER(Data[Date])&&Data[Purchase ID]<EARLIER(Data[Purchase ID])))<=3,and(CALCULATE(min(Data[Purchase ID]),FILTER(Data,Data[Customer]=EARLIER(Data[Customer])&&Data[Date]=EARLIER(Data[Date])&&Data[Purchase ID]>EARLIER(Data[Purchase ID])))-Data[Purchase ID]<=3,CALCULATE(min(Data[Purchase ID]),FILTER(Data,Data[Customer]=EARLIER(Data[Customer])&&Data[Date]=EARLIER(Data[Date])&&Data[Purchase ID]>EARLIER(Data[Purchase ID])))-Data[Purchase ID]>0)),"Yes","No"))
Hope this helps.
Hi @buttercream
Consider building a new Column in your table with below:
Perfect. Thank you !!!