Forum Discussion
Anonymous
2 years agoNot applicable
Convert DAX to Power M Query
First time trying to convert DAX to Power M Query and having a hard time getting started/finding exactly what I need in the help documents. Can someone help me convert this? And are there any resour...
- Anonymous2 years ago
Hi Anonymous
You can use group by in power query to count rows, you can refer to the folloing link.
Group rows of data (Power Query) - Microsoft Support
You can put the following code to advanced editor in power query to as a sample
let Source = Table.FromRows(Json.Document(Binary.Decompress(Binary.FromText("i45WMlTSUUpUitVBZyXBWclglhFc1gguawSRjQUA", BinaryEncoding.Base64), Compression.Deflate)), let _t = ((type nullable text) meta [Serialized.Text = true]) in type table [ID = _t, Type = _t]), #"Changed Type" = Table.TransformColumnTypes(Source,{{"ID", Int64.Type}, {"Type", type text}}), #"Grouped Rows" = Table.Group(#"Changed Type", {"ID", "Type"}, {{"Count", each Table.RowCount(_), Int64.Type}}) in #"Grouped Rows"Best Regards!
Yolo Zhu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
2 years agoNot applicable
Hi Anonymous
You can use group by in power query to count rows, you can refer to the folloing link.
Group rows of data (Power Query) - Microsoft Support
You can put the following code to advanced editor in power query to as a sample
let
Source = Table.FromRows(Json.Document(Binary.Decompress(Binary.FromText("i45WMlTSUUpUitVBZyXBWclglhFc1gguawSRjQUA", BinaryEncoding.Base64), Compression.Deflate)), let _t = ((type nullable text) meta [Serialized.Text = true]) in type table [ID = _t, Type = _t]),
#"Changed Type" = Table.TransformColumnTypes(Source,{{"ID", Int64.Type}, {"Type", type text}}),
#"Grouped Rows" = Table.Group(#"Changed Type", {"ID", "Type"}, {{"Count", each Table.RowCount(_), Int64.Type}})
in
#"Grouped Rows"
Best Regards!
Yolo Zhu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.