How to count distinct userid by each dayin Klipfolio?












0















I need to count the number of unique values and to group them by day.



For example, I have :



user id | Date



1 | 01.-11-2018



1 | 01-11-2018
2 | 02-11-2018



3 | 03-11-2018



1 | 03-11-2018



In result i need



01-11-2018 | 1



02-11-2018 | 1



03-11-2018 | 2



I need do it in Klipfolio










share|improve this question



























    0















    I need to count the number of unique values and to group them by day.



    For example, I have :



    user id | Date



    1 | 01.-11-2018



    1 | 01-11-2018
    2 | 02-11-2018



    3 | 03-11-2018



    1 | 03-11-2018



    In result i need



    01-11-2018 | 1



    02-11-2018 | 1



    03-11-2018 | 2



    I need do it in Klipfolio










    share|improve this question

























      0












      0








      0








      I need to count the number of unique values and to group them by day.



      For example, I have :



      user id | Date



      1 | 01.-11-2018



      1 | 01-11-2018
      2 | 02-11-2018



      3 | 03-11-2018



      1 | 03-11-2018



      In result i need



      01-11-2018 | 1



      02-11-2018 | 1



      03-11-2018 | 2



      I need do it in Klipfolio










      share|improve this question














      I need to count the number of unique values and to group them by day.



      For example, I have :



      user id | Date



      1 | 01.-11-2018



      1 | 01-11-2018
      2 | 02-11-2018



      3 | 03-11-2018



      1 | 03-11-2018



      In result i need



      01-11-2018 | 1



      02-11-2018 | 1



      03-11-2018 | 2



      I need do it in Klipfolio







      klipfolio






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 14 '18 at 9:02









      Софія ДичкоСофія Дичко

      11




      11
























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          You can use the GROUPBY() function with the aggregation in the method parameter set to Count Distinct. For example, if your dates are in column A of a spreadsheet and your values are in column B, your formula would look like this:



          GROUPBY(@A:A,@B:B,Count Distinct)





          share|improve this answer























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            1 Answer
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            0














            You can use the GROUPBY() function with the aggregation in the method parameter set to Count Distinct. For example, if your dates are in column A of a spreadsheet and your values are in column B, your formula would look like this:



            GROUPBY(@A:A,@B:B,Count Distinct)





            share|improve this answer




























              0














              You can use the GROUPBY() function with the aggregation in the method parameter set to Count Distinct. For example, if your dates are in column A of a spreadsheet and your values are in column B, your formula would look like this:



              GROUPBY(@A:A,@B:B,Count Distinct)





              share|improve this answer


























                0












                0








                0







                You can use the GROUPBY() function with the aggregation in the method parameter set to Count Distinct. For example, if your dates are in column A of a spreadsheet and your values are in column B, your formula would look like this:



                GROUPBY(@A:A,@B:B,Count Distinct)





                share|improve this answer













                You can use the GROUPBY() function with the aggregation in the method parameter set to Count Distinct. For example, if your dates are in column A of a spreadsheet and your values are in column B, your formula would look like this:



                GROUPBY(@A:A,@B:B,Count Distinct)






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 14 '18 at 17:36









                Adam DSAdam DS

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