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vividarinda's avatar
vividarinda
Helper II
2 years ago
Solved

Sort Latest Week Value

Hi All, How to sort the latest week values from the matrix below?   
  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi vividarinda ,

     

    (1)We can create a date table.

    Date = ADDCOLUMNS(CALENDAR(DATE(2024,1,1),DATE(2024,12,31)),"week",WEEKNUM([Date]))

    (2)We can create a column and a measure.

    ExtractedNumber = 
    VAR FirstSpace = SEARCH(" ", 'Table'[Week], 1)
    VAR SecondSpace = SEARCH(" ", 'Table'[Week], FirstSpace + 1)
    RETURN MID('Table'[Week], FirstSpace + 1, SecondSpace - FirstSpace - 1)
    Measure = var _max=MAXX(ALLSELECTED('Date'),[week])
    var _filter=MAXX(FILTER(ALLSELECTED('Table'),[Description] in VALUES('Table'[Description])&&[ExtractedNumber]=_max),[value])
    return _filter

    (3)Creating Model Relationship.

    (4)Sort - Turn off text wrap - Manually adjust column width to hide.

    Best Regards,

    Neeko Tang

    If this post  helps, then please consider Accept it as the solution  to help the other members find it more quickly. 

  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi vividarinda ,

     

    We can create a date table.

    Date = ADDCOLUMNS(CALENDAR(DATE(2024,1,1),DATE(2024,12,31)),"week",WEEKNUM([Date]),"month",MONTH([Date]))

    This is my test data. We can create columns.

    _month = 
    LEFT([month], FIND(" ", [month]) - 1)
    _month_num = MONTH( CONVERT([_month]&" "&1,DATETIME))

    We can create a measure.

    Measure 2 = 
    var _max=MAXX(ALLSELECTED('Date'),[month])
    var _filter=MAXX(FILTER(ALLSELECTED('Table (2)'),[Description] in VALUES('Table (2)'[Description])&&[_month_num]=_max),[value])
    return _filter

     

    If you still have problems following up, please create a new case, we suggest a case to solve only one problem, because it can get good help and give a better reference for other users!

     

    Best Regards,

    Neeko Tang

    If this post  helps, then please consider Accept it as the solution  to help the other members find it more quickly.