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Hello experts !
it is possible to do from this :
this
the important thing in this logic is that:
5 = A
3 = B
0 = C
in final table we have counts of that values for example in A if in CAT 1 we have 5 times value A in column CAT - A we will be have 5, etc
Solved! Go to Solution.
= Table.ReplaceValue(#"Changed Type",null,null,(x,y,z) => List.ReplaceMatchingItems({x},{{5,"A"},{3,"B"},{0,"C"}}){0},Table.ColumnNames(#"Changed Type"))
= Table.ReplaceValue(#"Changed Type",null,null,(x,y,z) => try {"A","B","C"}{List.PositionOf({5,3,0},x,0)} otherwise x,Table.ColumnNames(#"Changed Type"))
| Thanks to the great efforts by MS engineers to simplify syntax of DAX! Most beginners are SUCCESSFULLY MISLED to think that they could easily master DAX; but it turns out that the intricacy of the most frequently used RANKX() is still way beyond their comprehension! |
DAX is simple, but NOT EASY! |
= Table.ReplaceValue(#"Changed Type",null,null,(x,y,z) => List.ReplaceMatchingItems({x},{{5,"A"},{3,"B"},{0,"C"}}){0},Table.ColumnNames(#"Changed Type"))
= Table.ReplaceValue(#"Changed Type",null,null,(x,y,z) => try {"A","B","C"}{List.PositionOf({5,3,0},x,0)} otherwise x,Table.ColumnNames(#"Changed Type"))
| Thanks to the great efforts by MS engineers to simplify syntax of DAX! Most beginners are SUCCESSFULLY MISLED to think that they could easily master DAX; but it turns out that the intricacy of the most frequently used RANKX() is still way beyond their comprehension! |
DAX is simple, but NOT EASY! |
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