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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 a bunch of new columns. But is it possible to get a list of the values splitted by delimiter? So the output should be:
RowID Value
Row1 List = {a, b, c, d, a}
Row2 List = {a, c, c, c, c, a}
Row3 List = {a, c, f, h, j, r}
Thank you!
Solved! Go to Solution.
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"
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"
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