Power BI. Remove duplicates but keep null values
In Power BI, I need to remove duplicates of a column but keep the null values as they are 'pending'. Is there a way I can do it with DAX or the Query Editor?
powerbi dax powerquery m
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In Power BI, I need to remove duplicates of a column but keep the null values as they are 'pending'. Is there a way I can do it with DAX or the Query Editor?
powerbi dax powerquery m
add a comment |
In Power BI, I need to remove duplicates of a column but keep the null values as they are 'pending'. Is there a way I can do it with DAX or the Query Editor?
powerbi dax powerquery m
In Power BI, I need to remove duplicates of a column but keep the null values as they are 'pending'. Is there a way I can do it with DAX or the Query Editor?
powerbi dax powerquery m
powerbi dax powerquery m
edited Nov 15 '18 at 17:04
Alexis Olson
14.6k21834
14.6k21834
asked Nov 15 '18 at 8:04
deoosdeoos
174
174
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2 Answers
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Filter the table in two ways, without nulls and only nulls.
On the table without nulls, remove duplicates. Home > Remove Rows > Remove Duplicates
Append the null rows to this table.
The M code will look like this:
let
Source = <Data source or previous step reference here>,
AllNulls = Table.SelectRows(Source, each ([Column1] = null)),
NoNulls = Table.SelectRows(Source, each ([Column1] <> null)),
#"Removed Duplicates" = Table.Distinct(NoNulls),
#"Appended Query" = Table.Combine(#"Removed Duplicates", AllNulls)
in
#"Appended Query"
add a comment |
Try this in the Query Editor.
- Add an index column (Add Collumn tab > Index Column)
- Add a Custom Column with this formula ([Test] is your original column with nulls and duplicates.
- Rightclick the latest column [Temp] and select
Remove Duplicates
- Remove [Index] and [Temp] columns
The one thing you need to be careful about with this if your[Test]
column has a null in row N and the column also contains the value N in another row, then one of those rows gets removed when deleting duplicates.
– Alexis Olson
Nov 15 '18 at 17:07
@Alexis Olson. You're right off course. I made my solution more robust, by adding some random characters to the index value. See edit.
– Marco Vos
Nov 16 '18 at 8:27
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Filter the table in two ways, without nulls and only nulls.
On the table without nulls, remove duplicates. Home > Remove Rows > Remove Duplicates
Append the null rows to this table.
The M code will look like this:
let
Source = <Data source or previous step reference here>,
AllNulls = Table.SelectRows(Source, each ([Column1] = null)),
NoNulls = Table.SelectRows(Source, each ([Column1] <> null)),
#"Removed Duplicates" = Table.Distinct(NoNulls),
#"Appended Query" = Table.Combine(#"Removed Duplicates", AllNulls)
in
#"Appended Query"
add a comment |
Filter the table in two ways, without nulls and only nulls.
On the table without nulls, remove duplicates. Home > Remove Rows > Remove Duplicates
Append the null rows to this table.
The M code will look like this:
let
Source = <Data source or previous step reference here>,
AllNulls = Table.SelectRows(Source, each ([Column1] = null)),
NoNulls = Table.SelectRows(Source, each ([Column1] <> null)),
#"Removed Duplicates" = Table.Distinct(NoNulls),
#"Appended Query" = Table.Combine(#"Removed Duplicates", AllNulls)
in
#"Appended Query"
add a comment |
Filter the table in two ways, without nulls and only nulls.
On the table without nulls, remove duplicates. Home > Remove Rows > Remove Duplicates
Append the null rows to this table.
The M code will look like this:
let
Source = <Data source or previous step reference here>,
AllNulls = Table.SelectRows(Source, each ([Column1] = null)),
NoNulls = Table.SelectRows(Source, each ([Column1] <> null)),
#"Removed Duplicates" = Table.Distinct(NoNulls),
#"Appended Query" = Table.Combine(#"Removed Duplicates", AllNulls)
in
#"Appended Query"
Filter the table in two ways, without nulls and only nulls.
On the table without nulls, remove duplicates. Home > Remove Rows > Remove Duplicates
Append the null rows to this table.
The M code will look like this:
let
Source = <Data source or previous step reference here>,
AllNulls = Table.SelectRows(Source, each ([Column1] = null)),
NoNulls = Table.SelectRows(Source, each ([Column1] <> null)),
#"Removed Duplicates" = Table.Distinct(NoNulls),
#"Appended Query" = Table.Combine(#"Removed Duplicates", AllNulls)
in
#"Appended Query"
answered Nov 15 '18 at 17:04
Alexis OlsonAlexis Olson
14.6k21834
14.6k21834
add a comment |
add a comment |
Try this in the Query Editor.
- Add an index column (Add Collumn tab > Index Column)
- Add a Custom Column with this formula ([Test] is your original column with nulls and duplicates.
- Rightclick the latest column [Temp] and select
Remove Duplicates
- Remove [Index] and [Temp] columns
The one thing you need to be careful about with this if your[Test]
column has a null in row N and the column also contains the value N in another row, then one of those rows gets removed when deleting duplicates.
– Alexis Olson
Nov 15 '18 at 17:07
@Alexis Olson. You're right off course. I made my solution more robust, by adding some random characters to the index value. See edit.
– Marco Vos
Nov 16 '18 at 8:27
add a comment |
Try this in the Query Editor.
- Add an index column (Add Collumn tab > Index Column)
- Add a Custom Column with this formula ([Test] is your original column with nulls and duplicates.
- Rightclick the latest column [Temp] and select
Remove Duplicates
- Remove [Index] and [Temp] columns
The one thing you need to be careful about with this if your[Test]
column has a null in row N and the column also contains the value N in another row, then one of those rows gets removed when deleting duplicates.
– Alexis Olson
Nov 15 '18 at 17:07
@Alexis Olson. You're right off course. I made my solution more robust, by adding some random characters to the index value. See edit.
– Marco Vos
Nov 16 '18 at 8:27
add a comment |
Try this in the Query Editor.
- Add an index column (Add Collumn tab > Index Column)
- Add a Custom Column with this formula ([Test] is your original column with nulls and duplicates.
- Rightclick the latest column [Temp] and select
Remove Duplicates
- Remove [Index] and [Temp] columns
Try this in the Query Editor.
- Add an index column (Add Collumn tab > Index Column)
- Add a Custom Column with this formula ([Test] is your original column with nulls and duplicates.
- Rightclick the latest column [Temp] and select
Remove Duplicates
- Remove [Index] and [Temp] columns
edited Nov 16 '18 at 8:24
answered Nov 15 '18 at 15:01
Marco VosMarco Vos
1,998149
1,998149
The one thing you need to be careful about with this if your[Test]
column has a null in row N and the column also contains the value N in another row, then one of those rows gets removed when deleting duplicates.
– Alexis Olson
Nov 15 '18 at 17:07
@Alexis Olson. You're right off course. I made my solution more robust, by adding some random characters to the index value. See edit.
– Marco Vos
Nov 16 '18 at 8:27
add a comment |
The one thing you need to be careful about with this if your[Test]
column has a null in row N and the column also contains the value N in another row, then one of those rows gets removed when deleting duplicates.
– Alexis Olson
Nov 15 '18 at 17:07
@Alexis Olson. You're right off course. I made my solution more robust, by adding some random characters to the index value. See edit.
– Marco Vos
Nov 16 '18 at 8:27
The one thing you need to be careful about with this if your
[Test]
column has a null in row N and the column also contains the value N in another row, then one of those rows gets removed when deleting duplicates.– Alexis Olson
Nov 15 '18 at 17:07
The one thing you need to be careful about with this if your
[Test]
column has a null in row N and the column also contains the value N in another row, then one of those rows gets removed when deleting duplicates.– Alexis Olson
Nov 15 '18 at 17:07
@Alexis Olson. You're right off course. I made my solution more robust, by adding some random characters to the index value. See edit.
– Marco Vos
Nov 16 '18 at 8:27
@Alexis Olson. You're right off course. I made my solution more robust, by adding some random characters to the index value. See edit.
– Marco Vos
Nov 16 '18 at 8:27
add a comment |
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