How to get a DataFrame from the DataFrame with one column as a sum of values of other rows?



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0















I have a DataFrame in this way:



 shop_id item_price item_cnt_day day month year
59 9.00 1.0 02 01 2013
59 8.00 2.0 02 01 2013
25 10.00 4.0 05 02 2013
25 17.0 1.0 06 01 2013
25 10.00 1.0 15 01 2013


And I try to get the result like following DataFrame:



 shop_id all_revenue month year
59 25.00 01 2013
25 27.00 01 2013


I mean I want to get each shop's revenue in January 2013.



BUT, I don't know how to code in Pandas. Any help would be appreciated.










share|improve this question




























    0















    I have a DataFrame in this way:



     shop_id item_price item_cnt_day day month year
    59 9.00 1.0 02 01 2013
    59 8.00 2.0 02 01 2013
    25 10.00 4.0 05 02 2013
    25 17.0 1.0 06 01 2013
    25 10.00 1.0 15 01 2013


    And I try to get the result like following DataFrame:



     shop_id all_revenue month year
    59 25.00 01 2013
    25 27.00 01 2013


    I mean I want to get each shop's revenue in January 2013.



    BUT, I don't know how to code in Pandas. Any help would be appreciated.










    share|improve this question
























      0












      0








      0








      I have a DataFrame in this way:



       shop_id item_price item_cnt_day day month year
      59 9.00 1.0 02 01 2013
      59 8.00 2.0 02 01 2013
      25 10.00 4.0 05 02 2013
      25 17.0 1.0 06 01 2013
      25 10.00 1.0 15 01 2013


      And I try to get the result like following DataFrame:



       shop_id all_revenue month year
      59 25.00 01 2013
      25 27.00 01 2013


      I mean I want to get each shop's revenue in January 2013.



      BUT, I don't know how to code in Pandas. Any help would be appreciated.










      share|improve this question














      I have a DataFrame in this way:



       shop_id item_price item_cnt_day day month year
      59 9.00 1.0 02 01 2013
      59 8.00 2.0 02 01 2013
      25 10.00 4.0 05 02 2013
      25 17.0 1.0 06 01 2013
      25 10.00 1.0 15 01 2013


      And I try to get the result like following DataFrame:



       shop_id all_revenue month year
      59 25.00 01 2013
      25 27.00 01 2013


      I mean I want to get each shop's revenue in January 2013.



      BUT, I don't know how to code in Pandas. Any help would be appreciated.







      python pandas






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 15 '18 at 14:17









      FreAk PointFreAk Point

      798




      798






















          2 Answers
          2






          active

          oldest

          votes


















          4















          eval + groupby + sum



          You can assign a series via eval, then use groupby:



          res = df.eval('revenue=item_price * item_cnt_day')
          .groupby(['shop_id', 'month', 'year'], as_index=False)['revenue'].sum()


          You can, if you wish, query for January 2013 (before or after the above operations):



          res = res.query('month == 1 & year == 2013')

          print(res)

          shop_id month year revenue
          0 25 1 2013 27.0
          2 59 1 2013 25.0





          share|improve this answer
































            2














            I like filtering the dataframe first, to reduce number of unnecessary calculations:



            df.query('month == 1 and year == 2013')
            .assign(all_revenue = df.item_price * df.item_cnt_day)
            .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()


            Output:



             shop_id month year all_revenue
            0 25 1 2013 27.0
            1 59 1 2013 25.0


            Note: Because your column names are "friendly", no spaces nor special characters, you can use query method. If that doesn't work for your column naming then you need to use boolean indexing.



            df[(df['month'] == 1) & (df['year'] == 2013)]
            .assign(all_revenue = df.item_price * df.item_cnt_day)
            .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()





            share|improve this answer























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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4















              eval + groupby + sum



              You can assign a series via eval, then use groupby:



              res = df.eval('revenue=item_price * item_cnt_day')
              .groupby(['shop_id', 'month', 'year'], as_index=False)['revenue'].sum()


              You can, if you wish, query for January 2013 (before or after the above operations):



              res = res.query('month == 1 & year == 2013')

              print(res)

              shop_id month year revenue
              0 25 1 2013 27.0
              2 59 1 2013 25.0





              share|improve this answer





























                4















                eval + groupby + sum



                You can assign a series via eval, then use groupby:



                res = df.eval('revenue=item_price * item_cnt_day')
                .groupby(['shop_id', 'month', 'year'], as_index=False)['revenue'].sum()


                You can, if you wish, query for January 2013 (before or after the above operations):



                res = res.query('month == 1 & year == 2013')

                print(res)

                shop_id month year revenue
                0 25 1 2013 27.0
                2 59 1 2013 25.0





                share|improve this answer



























                  4












                  4








                  4








                  eval + groupby + sum



                  You can assign a series via eval, then use groupby:



                  res = df.eval('revenue=item_price * item_cnt_day')
                  .groupby(['shop_id', 'month', 'year'], as_index=False)['revenue'].sum()


                  You can, if you wish, query for January 2013 (before or after the above operations):



                  res = res.query('month == 1 & year == 2013')

                  print(res)

                  shop_id month year revenue
                  0 25 1 2013 27.0
                  2 59 1 2013 25.0





                  share|improve this answer
















                  eval + groupby + sum



                  You can assign a series via eval, then use groupby:



                  res = df.eval('revenue=item_price * item_cnt_day')
                  .groupby(['shop_id', 'month', 'year'], as_index=False)['revenue'].sum()


                  You can, if you wish, query for January 2013 (before or after the above operations):



                  res = res.query('month == 1 & year == 2013')

                  print(res)

                  shop_id month year revenue
                  0 25 1 2013 27.0
                  2 59 1 2013 25.0






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 15 '18 at 15:07

























                  answered Nov 15 '18 at 14:22









                  jppjpp

                  103k2167117




                  103k2167117























                      2














                      I like filtering the dataframe first, to reduce number of unnecessary calculations:



                      df.query('month == 1 and year == 2013')
                      .assign(all_revenue = df.item_price * df.item_cnt_day)
                      .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()


                      Output:



                       shop_id month year all_revenue
                      0 25 1 2013 27.0
                      1 59 1 2013 25.0


                      Note: Because your column names are "friendly", no spaces nor special characters, you can use query method. If that doesn't work for your column naming then you need to use boolean indexing.



                      df[(df['month'] == 1) & (df['year'] == 2013)]
                      .assign(all_revenue = df.item_price * df.item_cnt_day)
                      .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()





                      share|improve this answer



























                        2














                        I like filtering the dataframe first, to reduce number of unnecessary calculations:



                        df.query('month == 1 and year == 2013')
                        .assign(all_revenue = df.item_price * df.item_cnt_day)
                        .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()


                        Output:



                         shop_id month year all_revenue
                        0 25 1 2013 27.0
                        1 59 1 2013 25.0


                        Note: Because your column names are "friendly", no spaces nor special characters, you can use query method. If that doesn't work for your column naming then you need to use boolean indexing.



                        df[(df['month'] == 1) & (df['year'] == 2013)]
                        .assign(all_revenue = df.item_price * df.item_cnt_day)
                        .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()





                        share|improve this answer

























                          2












                          2








                          2







                          I like filtering the dataframe first, to reduce number of unnecessary calculations:



                          df.query('month == 1 and year == 2013')
                          .assign(all_revenue = df.item_price * df.item_cnt_day)
                          .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()


                          Output:



                           shop_id month year all_revenue
                          0 25 1 2013 27.0
                          1 59 1 2013 25.0


                          Note: Because your column names are "friendly", no spaces nor special characters, you can use query method. If that doesn't work for your column naming then you need to use boolean indexing.



                          df[(df['month'] == 1) & (df['year'] == 2013)]
                          .assign(all_revenue = df.item_price * df.item_cnt_day)
                          .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()





                          share|improve this answer













                          I like filtering the dataframe first, to reduce number of unnecessary calculations:



                          df.query('month == 1 and year == 2013')
                          .assign(all_revenue = df.item_price * df.item_cnt_day)
                          .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()


                          Output:



                           shop_id month year all_revenue
                          0 25 1 2013 27.0
                          1 59 1 2013 25.0


                          Note: Because your column names are "friendly", no spaces nor special characters, you can use query method. If that doesn't work for your column naming then you need to use boolean indexing.



                          df[(df['month'] == 1) & (df['year'] == 2013)]
                          .assign(all_revenue = df.item_price * df.item_cnt_day)
                          .groupby(['shop_id','month','year'], as_index=False)['all_revenue'].sum()






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 15 '18 at 14:38









                          Scott BostonScott Boston

                          58.7k73258




                          58.7k73258



























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