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The idea is to find the min of the value column related to the department column, but the min has to be different from 0.
then we want the second and third min
Here are the excel function used:
Small 1 = MINIFS($C:$C;$B:$B;$B2;$C:$C;"<>0")
Small 2 =MINIFS($C:$C;$B:$B;$B2;$C:$C;"<>0";$C:$C;CONCAT("<>";$D2))
Small 3 = =MINIFS($C:$C;$B:$B;$B2;$C:$C;"<>0";$C:$C;CONCAT("<>";$D2);$C:$C;CONCAT("<>";$E2))
I have put some data to illustrate:
Country | Departement | value | Min 1 | Min 2 | Min 3 |
CH | a | 0,824831 | 0,491241 | 0,759035 | 0,824831 |
CH | b | 0 | 0,562287 | 0,812621 | 0,866854 |
CH | c | 0,02241 | 0,02241 | 0,09508 | 0,195738 |
CH | d | 0,032302 | 0,032302 | 0,57665 | 0,748025 |
CH | e | 0,411674 | 0,411674 | 0,710219 | 0,729571 |
CH | f | 0,311402 | 0,131588 | 0,233024 | 0,311402 |
ES | a | 0,759035 | 0,491241 | 0,759035 | 0,824831 |
ES | b | 0,866854 | 0,562287 | 0,812621 | 0,866854 |
ES | c | 0 | 0,02241 | 0,09508 | 0,195738 |
ES | d | 0,760727 | 0,032302 | 0,57665 | 0,748025 |
ES | e | 0,729571 | 0,411674 | 0,710219 | 0,729571 |
ES | f | 0,233024 | 0,131588 | 0,233024 | 0,311402 |
BR | a | 0 | 0,491241 | 0,759035 | 0,824831 |
BR | b | 0,812621 | 0,562287 | 0,812621 | 0,866854 |
BR | c | 0,09508 | 0,02241 | 0,09508 | 0,195738 |
BR | d | 0,57665 | 0,032302 | 0,57665 | 0,748025 |
BR | e | 0,796108 | 0,411674 | 0,710219 | 0,729571 |
BR | f | 0,520732 | 0,131588 | 0,233024 | 0,311402 |
PR | a | 0,491241 | 0,491241 | 0,759035 | 0,824831 |
PR | b | 0,562287 | 0,562287 | 0,812621 | 0,866854 |
PR | c | 0,195738 | 0,02241 | 0,09508 | 0,195738 |
PR | d | 0,748025 | 0,032302 | 0,57665 | 0,748025 |
PR | e | 0,710219 | 0,411674 | 0,710219 | 0,729571 |
PR | f | 0,131588 | 0,131588 | 0,233024 | 0,311402 |
Solved! Go to Solution.
Hi @AlexMunday
Create a calculated column for the first min:
Min1 = CALCULATE ( MIN ( Table1[value ] ); ALLEXCEPT ( Table1; Table1[Departement ] ); Table1[value] <> 0 )
Then for min2 using the newly created column:
Min2 = CALCULATE ( MIN ( Table1[value ] ); ALLEXCEPT ( Table1; Table1[Departement ] ); Table1[value] <> 0; Table1[value] <> EARLIER( Table1[Min1]) )
and then for min3
Min3 = CALCULATE ( MIN ( Table1[value ] ); ALLEXCEPT ( Table1; Table1[Departement ] ); Table1[value] <> 0; Table1[value] <> EARLIER( Table1[Min1]);
Table1[value] <> EARLIER( Table1[Min2]) )
Hi @AlexMunday
Create a calculated column for the first min:
Min1 = CALCULATE ( MIN ( Table1[value ] ); ALLEXCEPT ( Table1; Table1[Departement ] ); Table1[value] <> 0 )
Then for min2 using the newly created column:
Min2 = CALCULATE ( MIN ( Table1[value ] ); ALLEXCEPT ( Table1; Table1[Departement ] ); Table1[value] <> 0; Table1[value] <> EARLIER( Table1[Min1]) )
and then for min3
Min3 = CALCULATE ( MIN ( Table1[value ] ); ALLEXCEPT ( Table1; Table1[Departement ] ); Table1[value] <> 0; Table1[value] <> EARLIER( Table1[Min1]);
Table1[value] <> EARLIER( Table1[Min2]) )
Thank you it's working perfectly
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