Return rows that has 3 or more consecutive values as 1 with Order by on Date & Group By Region












0















I tried Tabibitosan method but no help. Can you please suggest some other solution for below scenario.



Using Oracle 11g:



Please find the below table format ,



Region  Date      Value
East 1/1/2018 1
East 1/2/2018 1
East 1/3/2018 0
East 1/4/2018 1
East 1/5/2018 1
East 1/7/2018 1
West 1/9/2018 0
West 1/10/2018 0
West 2/3/2018 1
West 2/4/2018 1
East 2/5/2018 1
West 2/8/2018 0
West 2/9/2018 0
West 2/10/2018 0


From the above table I should return the rows that has value 1 occurred 3 or more times with respect date(order by) and Region.



**My Output:**

Region Date Value
East 1/4/2018 1
East 1/5/2018 1
East 1/7/2018 1


Note : The date in the Date column may not contain all the days. Say, in the above 1/6/2018 is missing which is fine. I need to look for 'Value' column that has 1 consecutive for 3 or more rows when ordered by 'Date'.










share|improve this question





























    0















    I tried Tabibitosan method but no help. Can you please suggest some other solution for below scenario.



    Using Oracle 11g:



    Please find the below table format ,



    Region  Date      Value
    East 1/1/2018 1
    East 1/2/2018 1
    East 1/3/2018 0
    East 1/4/2018 1
    East 1/5/2018 1
    East 1/7/2018 1
    West 1/9/2018 0
    West 1/10/2018 0
    West 2/3/2018 1
    West 2/4/2018 1
    East 2/5/2018 1
    West 2/8/2018 0
    West 2/9/2018 0
    West 2/10/2018 0


    From the above table I should return the rows that has value 1 occurred 3 or more times with respect date(order by) and Region.



    **My Output:**

    Region Date Value
    East 1/4/2018 1
    East 1/5/2018 1
    East 1/7/2018 1


    Note : The date in the Date column may not contain all the days. Say, in the above 1/6/2018 is missing which is fine. I need to look for 'Value' column that has 1 consecutive for 3 or more rows when ordered by 'Date'.










    share|improve this question



























      0












      0








      0


      1






      I tried Tabibitosan method but no help. Can you please suggest some other solution for below scenario.



      Using Oracle 11g:



      Please find the below table format ,



      Region  Date      Value
      East 1/1/2018 1
      East 1/2/2018 1
      East 1/3/2018 0
      East 1/4/2018 1
      East 1/5/2018 1
      East 1/7/2018 1
      West 1/9/2018 0
      West 1/10/2018 0
      West 2/3/2018 1
      West 2/4/2018 1
      East 2/5/2018 1
      West 2/8/2018 0
      West 2/9/2018 0
      West 2/10/2018 0


      From the above table I should return the rows that has value 1 occurred 3 or more times with respect date(order by) and Region.



      **My Output:**

      Region Date Value
      East 1/4/2018 1
      East 1/5/2018 1
      East 1/7/2018 1


      Note : The date in the Date column may not contain all the days. Say, in the above 1/6/2018 is missing which is fine. I need to look for 'Value' column that has 1 consecutive for 3 or more rows when ordered by 'Date'.










      share|improve this question
















      I tried Tabibitosan method but no help. Can you please suggest some other solution for below scenario.



      Using Oracle 11g:



      Please find the below table format ,



      Region  Date      Value
      East 1/1/2018 1
      East 1/2/2018 1
      East 1/3/2018 0
      East 1/4/2018 1
      East 1/5/2018 1
      East 1/7/2018 1
      West 1/9/2018 0
      West 1/10/2018 0
      West 2/3/2018 1
      West 2/4/2018 1
      East 2/5/2018 1
      West 2/8/2018 0
      West 2/9/2018 0
      West 2/10/2018 0


      From the above table I should return the rows that has value 1 occurred 3 or more times with respect date(order by) and Region.



      **My Output:**

      Region Date Value
      East 1/4/2018 1
      East 1/5/2018 1
      East 1/7/2018 1


      Note : The date in the Date column may not contain all the days. Say, in the above 1/6/2018 is missing which is fine. I need to look for 'Value' column that has 1 consecutive for 3 or more rows when ordered by 'Date'.







      sql oracle oracle11g analytics






      share|improve this question















      share|improve this question













      share|improve this question




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      edited Nov 23 '18 at 16:09







      ABR

















      asked Nov 23 '18 at 15:56









      ABRABR

      274




      274
























          2 Answers
          2






          active

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          0














          Try the following if you wish to get all the rows which match the condition.



          with data
          as (
          select x.region,x.date1,x.value,x.pattern_start,x.rnk
          from (
          select region
          ,date1
          ,value
          ,row_number() over(order by region,date1) as rnk
          ,case when value=1
          and lead(value,1) over(partition by region order by date1) = 1
          and lead(value,2) over(partition by region order by date1) = 1
          then row_number() over(order by region,date1)
          end as pattern_start
          ,lead(value,2) over(partition by region order by date1) as next_val_2
          ,lead(value,3) over(partition by region order by date1) as next_val_3
          from t)x
          )
          select *
          from data y
          where y.rnk in (select pattern_start from data union all
          select pattern_start+1 from data union all
          select pattern_start+2 from data
          )
          order by 1,2


          Demo Link
          https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=1aa6d5de2b0ec375f659d0243aba350a






          share|improve this answer

































            1














            Just use lead():



            select t.*
            from (select t.*,
            lead(value) over (partition by region order by date) as value_1,
            lead(value, 2) over (partition by region order by date) as value_2
            from t
            ) t
            where value = 1 and value_1 = 1 and value_2 = 1;


            In the event that you have 4 or more in a row and only want the first one, you can add a lag():



            select t.*
            from (select t.*,
            lag(value) over (partition by region order by date) as prev_value,
            lead(value) over (partition by region order by date) as value_1,
            lead(value, 2) over (partition by region order by date) as value_2
            from t
            ) t
            where value = 1 and value_1 = 1 and value_2 = 1 and
            (prev_value is null or prev_value <> 1);





            share|improve this answer
























            • Thank you so much I'm verifying your query. Will the first query return all the rows when there is consecutive 4 values?

              – ABR
              Nov 23 '18 at 16:39











            • @ABR . . . It would return the first two rows, because each begin a sequence of 3.

              – Gordon Linoff
              Nov 23 '18 at 16:42











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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

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            active

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            0














            Try the following if you wish to get all the rows which match the condition.



            with data
            as (
            select x.region,x.date1,x.value,x.pattern_start,x.rnk
            from (
            select region
            ,date1
            ,value
            ,row_number() over(order by region,date1) as rnk
            ,case when value=1
            and lead(value,1) over(partition by region order by date1) = 1
            and lead(value,2) over(partition by region order by date1) = 1
            then row_number() over(order by region,date1)
            end as pattern_start
            ,lead(value,2) over(partition by region order by date1) as next_val_2
            ,lead(value,3) over(partition by region order by date1) as next_val_3
            from t)x
            )
            select *
            from data y
            where y.rnk in (select pattern_start from data union all
            select pattern_start+1 from data union all
            select pattern_start+2 from data
            )
            order by 1,2


            Demo Link
            https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=1aa6d5de2b0ec375f659d0243aba350a






            share|improve this answer






























              0














              Try the following if you wish to get all the rows which match the condition.



              with data
              as (
              select x.region,x.date1,x.value,x.pattern_start,x.rnk
              from (
              select region
              ,date1
              ,value
              ,row_number() over(order by region,date1) as rnk
              ,case when value=1
              and lead(value,1) over(partition by region order by date1) = 1
              and lead(value,2) over(partition by region order by date1) = 1
              then row_number() over(order by region,date1)
              end as pattern_start
              ,lead(value,2) over(partition by region order by date1) as next_val_2
              ,lead(value,3) over(partition by region order by date1) as next_val_3
              from t)x
              )
              select *
              from data y
              where y.rnk in (select pattern_start from data union all
              select pattern_start+1 from data union all
              select pattern_start+2 from data
              )
              order by 1,2


              Demo Link
              https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=1aa6d5de2b0ec375f659d0243aba350a






              share|improve this answer




























                0












                0








                0







                Try the following if you wish to get all the rows which match the condition.



                with data
                as (
                select x.region,x.date1,x.value,x.pattern_start,x.rnk
                from (
                select region
                ,date1
                ,value
                ,row_number() over(order by region,date1) as rnk
                ,case when value=1
                and lead(value,1) over(partition by region order by date1) = 1
                and lead(value,2) over(partition by region order by date1) = 1
                then row_number() over(order by region,date1)
                end as pattern_start
                ,lead(value,2) over(partition by region order by date1) as next_val_2
                ,lead(value,3) over(partition by region order by date1) as next_val_3
                from t)x
                )
                select *
                from data y
                where y.rnk in (select pattern_start from data union all
                select pattern_start+1 from data union all
                select pattern_start+2 from data
                )
                order by 1,2


                Demo Link
                https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=1aa6d5de2b0ec375f659d0243aba350a






                share|improve this answer















                Try the following if you wish to get all the rows which match the condition.



                with data
                as (
                select x.region,x.date1,x.value,x.pattern_start,x.rnk
                from (
                select region
                ,date1
                ,value
                ,row_number() over(order by region,date1) as rnk
                ,case when value=1
                and lead(value,1) over(partition by region order by date1) = 1
                and lead(value,2) over(partition by region order by date1) = 1
                then row_number() over(order by region,date1)
                end as pattern_start
                ,lead(value,2) over(partition by region order by date1) as next_val_2
                ,lead(value,3) over(partition by region order by date1) as next_val_3
                from t)x
                )
                select *
                from data y
                where y.rnk in (select pattern_start from data union all
                select pattern_start+1 from data union all
                select pattern_start+2 from data
                )
                order by 1,2


                Demo Link
                https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=1aa6d5de2b0ec375f659d0243aba350a







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 23 '18 at 17:06

























                answered Nov 23 '18 at 16:46









                George JosephGeorge Joseph

                1,44249




                1,44249

























                    1














                    Just use lead():



                    select t.*
                    from (select t.*,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1;


                    In the event that you have 4 or more in a row and only want the first one, you can add a lag():



                    select t.*
                    from (select t.*,
                    lag(value) over (partition by region order by date) as prev_value,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1 and
                    (prev_value is null or prev_value <> 1);





                    share|improve this answer
























                    • Thank you so much I'm verifying your query. Will the first query return all the rows when there is consecutive 4 values?

                      – ABR
                      Nov 23 '18 at 16:39











                    • @ABR . . . It would return the first two rows, because each begin a sequence of 3.

                      – Gordon Linoff
                      Nov 23 '18 at 16:42
















                    1














                    Just use lead():



                    select t.*
                    from (select t.*,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1;


                    In the event that you have 4 or more in a row and only want the first one, you can add a lag():



                    select t.*
                    from (select t.*,
                    lag(value) over (partition by region order by date) as prev_value,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1 and
                    (prev_value is null or prev_value <> 1);





                    share|improve this answer
























                    • Thank you so much I'm verifying your query. Will the first query return all the rows when there is consecutive 4 values?

                      – ABR
                      Nov 23 '18 at 16:39











                    • @ABR . . . It would return the first two rows, because each begin a sequence of 3.

                      – Gordon Linoff
                      Nov 23 '18 at 16:42














                    1












                    1








                    1







                    Just use lead():



                    select t.*
                    from (select t.*,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1;


                    In the event that you have 4 or more in a row and only want the first one, you can add a lag():



                    select t.*
                    from (select t.*,
                    lag(value) over (partition by region order by date) as prev_value,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1 and
                    (prev_value is null or prev_value <> 1);





                    share|improve this answer













                    Just use lead():



                    select t.*
                    from (select t.*,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1;


                    In the event that you have 4 or more in a row and only want the first one, you can add a lag():



                    select t.*
                    from (select t.*,
                    lag(value) over (partition by region order by date) as prev_value,
                    lead(value) over (partition by region order by date) as value_1,
                    lead(value, 2) over (partition by region order by date) as value_2
                    from t
                    ) t
                    where value = 1 and value_1 = 1 and value_2 = 1 and
                    (prev_value is null or prev_value <> 1);






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 23 '18 at 16:08









                    Gordon LinoffGordon Linoff

                    764k35296400




                    764k35296400













                    • Thank you so much I'm verifying your query. Will the first query return all the rows when there is consecutive 4 values?

                      – ABR
                      Nov 23 '18 at 16:39











                    • @ABR . . . It would return the first two rows, because each begin a sequence of 3.

                      – Gordon Linoff
                      Nov 23 '18 at 16:42



















                    • Thank you so much I'm verifying your query. Will the first query return all the rows when there is consecutive 4 values?

                      – ABR
                      Nov 23 '18 at 16:39











                    • @ABR . . . It would return the first two rows, because each begin a sequence of 3.

                      – Gordon Linoff
                      Nov 23 '18 at 16:42

















                    Thank you so much I'm verifying your query. Will the first query return all the rows when there is consecutive 4 values?

                    – ABR
                    Nov 23 '18 at 16:39





                    Thank you so much I'm verifying your query. Will the first query return all the rows when there is consecutive 4 values?

                    – ABR
                    Nov 23 '18 at 16:39













                    @ABR . . . It would return the first two rows, because each begin a sequence of 3.

                    – Gordon Linoff
                    Nov 23 '18 at 16:42





                    @ABR . . . It would return the first two rows, because each begin a sequence of 3.

                    – Gordon Linoff
                    Nov 23 '18 at 16:42


















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