Forum Discussion
Overlap table
- Anonymous7 years ago
Than you for your help! However this is just a subset. Doing this for the whole dataset would take quite a lot of time, because there are many combinations possible.
In another topic I found this:
This did not bring me to the exact table I wanted, so I did the rest by hand. However this took quite a lot of hand work, because all all possible orderings of products appeared in different cells (a,b,c - a,c,b - b,a,c - etc.).
The 5 upper left means that there are in total 5 zipcodes with product A.
The 2 upper middle means that there are in total 2 zipcodes with as well product A as product B.
The upper right means that there are no zipcodes with as well product A as product C.
I'm mainly interested in the overlap (how many people (zipcodes) have more than 1 product and which product combinations occur most often), so the values for A-A, B-B and C-C are not that important to me. It's mainly about the rest of the values in the matrix.
Anonymous ,
So your requirement is to achieve the distinct count of zipcode with three combinations "A&B", "B&C" and "A&C", right? You can create three calculate columns and three measures using DAX below:
Calculate columns:
Combination A&B = CALCULATE(CONCATENATEX('Table', 'Table'[Product], "&"), FILTER(ALLEXCEPT('Table', 'Table'[Zip Code]), 'Table'[Product] = "A" || 'Table'[Product] = "B"))
Combination B&C = CALCULATE(CONCATENATEX('Table', 'Table'[Product], "&"), FILTER(ALLEXCEPT('Table', 'Table'[Zip Code]), 'Table'[Product] = "B" || 'Table'[Product] = "C"))
Combination A&C = CALCULATE(CONCATENATEX('Table', 'Table'[Product], "&"), FILTER(ALLEXCEPT('Table', 'Table'[Zip Code]), 'Table'[Product] = "A" || 'Table'[Product] = "C"))
Measures:
Count A&B = CALCULATE(DISTINCTCOUNT('Table'[Zip Code]), FILTER('Table', 'Table'[Combination A&B] = "A&B"))
Count B&C = CALCULATE(DISTINCTCOUNT('Table'[Zip Code]), FILTER('Table', 'Table'[Combination B&C] = "B&C"))
Count A&C = CALCULATE(DISTINCTCOUNT('Table'[Zip Code]), FILTER('Table', 'Table'[Combination A&C] = "A&C"))
The result will be like below:
Community Support Team _ Jimmy Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
- Anonymous7 years agoNot applicable
Than you for your help! However this is just a subset. Doing this for the whole dataset would take quite a lot of time, because there are many combinations possible.
In another topic I found this:
This did not bring me to the exact table I wanted, so I did the rest by hand. However this took quite a lot of hand work, because all all possible orderings of products appeared in different cells (a,b,c - a,c,b - b,a,c - etc.).