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Igorpole's avatar
Igorpole
Icon for Helper I rankHelper I
5 years ago
Solved

Column values to list (by delimiter)

Hi!   Let's say I have such a table:   RowID Value Row1 a,b,c,d,a Row2 a,c,c,c,c,a Row3 a,c,f,h,j,r   It's very easy using UI split a "Value" column by delimiter (comma in this case) and get...
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  • Anonymous's avatar
    Anonymous
    5 years ago

    try this

     

    let
    Origine = Table.FromRows(Json.Document(Binary.Decompress(Binary.FromText("i45WCsovN1TSUUrUSdJJ1knRSVSK1QELGoEFk6EQLmwMFU7TydDJ0ilSio0FAA==", BinaryEncoding.Base64), Compression.Deflate)), let _t = ((type nullable text) meta [Serialized.Text = true]) in type table [RowID = _t, Value = _t]),
    #"Modificato tipo" = Table.TransformColumnTypes(Origine,{{"RowID", type text}, {"Value", type text}}),
    #"Aggiunta colonna personalizzata" = Table.AddColumn(#"Modificato tipo", "list_Values", each Text.Split([Value],","))
in
    #"Aggiunta colonna personalizzata"
Anonymous's avatar
Anonymous
Not applicable
5 years ago

try this

 

let
    Origine = Table.FromRows(Json.Document(Binary.Decompress(Binary.FromText("i45WCsovN1TSUUrUSdJJ1knRSVSK1QELGoEFk6EQLmwMFU7TydDJ0ilSio0FAA==", BinaryEncoding.Base64), Compression.Deflate)), let _t = ((type nullable text) meta [Serialized.Text = true]) in type table [RowID = _t, Value = _t]),
    #"Modificato tipo" = Table.TransformColumnTypes(Origine,{{"RowID", type text}, {"Value", type text}}),
    #"Aggiunta colonna personalizzata" = Table.AddColumn(#"Modificato tipo", "list_Values", each Text.Split([Value],","))
in
    #"Aggiunta colonna personalizzata"