Forum Discussion
Dynamically removing columns ending with even numbers
Hi All,
In Power Query, I have split a column based on "," delimiter and the result produced one column split into 10 columns and columns names have been named as "Column_1", "Column_2", and till "Column_10". Now today based on the values I get 10 columns but tomorrow I may get 20 columns. So, after split, I need to remove all the columns which is ending with even numbers like "Column_2", "Column_4" till the last even numbered columns. Is there a way to do it?
Regards,
Mohanraj
Hi
With 10 columns
= Table.SplitColumn(#"Changed Type 1", "Details", each List.Alternate(Text.Split(_,","),1,1,1), 10)
or
= Table.SplitColumn(
#"Changed Type 1",
"Details",
each List.Alternate(Text.Split(_,","),1,1,1),
List.Max(
Table.TransformColumns(
#"Changed Type 1",
{{"Details", each Number.IntegerDivide(Text.Length(Text.Select(_,",")) + 1, 2) + 1, Int64.Type}}
)[Details]
)
)https://goodly.co.in/split-by-variable-columns-in-power-query/
Stéphane
6 Replies
Hi
With 10 columns
= Table.SplitColumn(#"Changed Type 1", "Details", each List.Alternate(Text.Split(_,","),1,1,1), 10)
or
= Table.SplitColumn(
#"Changed Type 1",
"Details",
each List.Alternate(Text.Split(_,","),1,1,1),
List.Max(
Table.TransformColumns(
#"Changed Type 1",
{{"Details", each Number.IntegerDivide(Text.Length(Text.Select(_,",")) + 1, 2) + 1, Int64.Type}}
)[Details]
)
)https://goodly.co.in/split-by-variable-columns-in-power-query/
Stéphane
Thanks you very much slorin your solution worked like a charm.
Hi
= Table.SplitColumn(Your_Source, "Your_Column", each List.Alternate(Text.Split(_,","),1,1,1))
Stéphane
Hi slorin ,
Below is the screenshot of my Power Query and Advance editor. In the post I have mentioned the column name having "_", but in reality its ".". So, I tried to replace "_" to "." in the advance editor, but it didnt work.
Let me know as what is wrong.
The principle is not to delete even columns but to select only odd columns with the List.Alternate
= Table.SplitColumn(#"Changed Type 1", "Details", each List.Alternate(Text.Split(_,","),1,1,1))
Stéphane
Hi slorin your code worked partially. Meaning the code worked to the lowest length of the column values.
Eg: For column value 1, raw, 3 I got the result as 1, 3 whereas for the column value 2, name, house, 10, 6, 20 I got the result as 2, house.
