Can I create a segment with an aggregate using count between in AEP

[an1989]

Hi,

I am trying to create a segment with audience of count of Unique purchases (SKU's) between 2 and 5 in the last 10 days.

Can I do this in one single segment in AEP?

Thanks

[Danny-Miller]

@an1989 A few things to remember:

1. Segments do not produce an "audience count".  They are rules that define if someone qualifies for membership in a Segment.
2. You can do a Count of Purchases (not Count Distinct) using Aggregate functions in the UI. https://experienceleague.adobe.com/docs/experience-platform/segmentation/ui/segment-builder.html?lan...
3. You can do a Count Distinct in PQL.  https://experienceleague.adobe.com/docs/experience-platform/segmentation/pql/array-functions.html?la...
4. Even though you can do a Count Distinct, this is within the context of a single Array.  I do not think this will do a count Distinct across Events, but it is worth a try.

[an1989]

Thanks @Danny-Miller  for the response.

Sorry for not being clear with my question above.  I am trying to find the qualified audience who made more than 2 purchases and less than 5 purchases in last 10 days.

I earlier tried using the aggregate function count, but that doesn't have between option. I can either select greater than or equal to or less than equal to. But not both in the same segment. Is there any alternative approach here?

If I do count distinct of product(sku) purchases in PQL, Can I get the qualified audience here?

Thanks

[an1989]

Hi @Danny-Miller , just checking in to see if you have some insight on this?

Thanks

[Danny-Miller · Accepted solution]

@an1989 I built this in the UI.

Here is the PQL:

CHAIN(xEvent, timestamp, [C0: WHAT(true) WHEN(<= 10 days before now) COUNT(ALL) WHERE(C0[productListItems.exists(SKU.isNotNull())].count() >= 2 and C0[productListItems.exists(SKU.isNotNull())].count() <= 5)])

[an1989]

Thanks @Danny-Miller . It worked

Is there a way we can accommodate the "distinct"  for SKU's ? 

[Danny-Miller]

Not sure.  Did you try PQL?

distinct().count()