Calculating possible values

[snehamahaveer]

Hi All,

I've been trying to solve this mystery for a while now.

I have a list of customers with their proposed Discount code and their discount codes used at transaction. So my end goal is to find the number of customers by month who have been using the proposed discount code (among multiple transaction discount codes if used).

sum(aggr(if(ProposedDiscCode= Concat(DISTINCT Transaction_Discount_Code, ','),1,0),Customer,Month,Store))

The above formula helps me identify the number but when the customer uses only the proposed discount code and none others.

Customer	ProposedDiscCode	Transaction_Discount_Code
Adam	IG	IG,IH,IL
Sneha	IH	IH
Tim	IG	IG
Joe	IF	IF,IG,IY

So my formula for the above table of values gives the count as 2, whereas I want 4.

Can you please help me with this.

Thanks in advance,

S

[sunny_talwar]

May be try this

Sum(If(SubStringCount(Transaction_Discount_Code, ProposedDiscCode) > 0, 1, 0))

[sunny_talwar]

or this

Sum(If(Index(Transaction_Discount_Code, ProposedDiscCode), 1, 0))

[snehamahaveer]

Hi Sunny,

Can the expression be enabled to look at distinct transaction discount codes used? because in a month they may have multiple transactions and your formula calculates it at that level.

Thanks

S

[sunny_talwar]

I am not sure I understand... but may be this

Sum(Aggr(If(SubStringCount(Transaction_Discount_Code, ProposedDiscCode) > 0, 1, 0), Customer, Month, Store))

[snehamahaveer]

Hi Sunny,

It gives the same value as my expression above.

[sunny_talwar]

Would you be able to provide a sample where you have Month and Store with expected output from the sample provided?

[snehamahaveer]

Hi Sunny,

I have given a sample data.

My formula gives the total correct disc code customers to be 5, but the actual correct discount code customers should be 9.

[sunny_talwar · Accepted solution]

If this is how your data is, then try this

=Count(DISTINCT If(Transaction_Discount_Code = ProposedDiscCode, Customer))

[snehamahaveer]

Hi Sunny,

This formula works.:) many thanks for helping me so quick.!

However, I fail to understand the logic behind this, can you please help me understand it?

Regards,

S