pandas match date in one df with timeframe in another, then groupby-sum










1















I have two dataframes, test1 and test2. For each ID value in test2, I want to check the date in test2 and compare it to the date ranges for that same ID value in test1. If any of the date's in test2 are within a date range in test1, sum the amount column and assign that sum as an additional column in test1.



Output:



So the new test1 df will have a column amount_sum which is the sum of all amounts in test2 where the date is within the date range of test1 - for that ID



import random
import string

test1 = pd.DataFrame(
'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
'date1':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)],
'date2':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(101, 200, 1))) for n in range(100)]
)

test2 = pd.DataFrame(
'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
'amount':[random.choice([1,2,3,5,10]) for n in range(100)],
'date':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)]
)









share|improve this question




























    1















    I have two dataframes, test1 and test2. For each ID value in test2, I want to check the date in test2 and compare it to the date ranges for that same ID value in test1. If any of the date's in test2 are within a date range in test1, sum the amount column and assign that sum as an additional column in test1.



    Output:



    So the new test1 df will have a column amount_sum which is the sum of all amounts in test2 where the date is within the date range of test1 - for that ID



    import random
    import string

    test1 = pd.DataFrame(
    'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
    'date1':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)],
    'date2':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(101, 200, 1))) for n in range(100)]
    )

    test2 = pd.DataFrame(
    'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
    'amount':[random.choice([1,2,3,5,10]) for n in range(100)],
    'date':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)]
    )









    share|improve this question


























      1












      1








      1








      I have two dataframes, test1 and test2. For each ID value in test2, I want to check the date in test2 and compare it to the date ranges for that same ID value in test1. If any of the date's in test2 are within a date range in test1, sum the amount column and assign that sum as an additional column in test1.



      Output:



      So the new test1 df will have a column amount_sum which is the sum of all amounts in test2 where the date is within the date range of test1 - for that ID



      import random
      import string

      test1 = pd.DataFrame(
      'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
      'date1':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)],
      'date2':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(101, 200, 1))) for n in range(100)]
      )

      test2 = pd.DataFrame(
      'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
      'amount':[random.choice([1,2,3,5,10]) for n in range(100)],
      'date':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)]
      )









      share|improve this question
















      I have two dataframes, test1 and test2. For each ID value in test2, I want to check the date in test2 and compare it to the date ranges for that same ID value in test1. If any of the date's in test2 are within a date range in test1, sum the amount column and assign that sum as an additional column in test1.



      Output:



      So the new test1 df will have a column amount_sum which is the sum of all amounts in test2 where the date is within the date range of test1 - for that ID



      import random
      import string

      test1 = pd.DataFrame(
      'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
      'date1':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)],
      'date2':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(101, 200, 1))) for n in range(100)]
      )

      test2 = pd.DataFrame(
      'ID':[''.join(random.choice(string.ascii_letters[0:4]) for _ in range(3)) for n in range(100)],
      'amount':[random.choice([1,2,3,5,10]) for n in range(100)],
      'date':[pd.to_datetime(random.choice(['01-01-2018','05-01-2018','06-01-2018','08-01-2018','09-01-2018'])) + pd.DateOffset(int(np.random.randint(0, 100, 1))) for n in range(100)]
      )






      python pandas






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      edited Nov 13 '18 at 6:03







      jchaykow

















      asked Nov 13 '18 at 5:55









      jchaykowjchaykow

      540320




      540320






















          1 Answer
          1






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          3














          Use:



          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])


          Sample:



          #https://stackoverflow.com/q/21494489
          np.random.seed(123)

          #https://stackoverflow.com/a/50559321/2901002
          def gen(start, end, n):
          start_u = start.value//10**9
          end_u = end.value//10**9

          return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')

          n = 10
          test1 = pd.DataFrame(
          'ID':np.random.choice(list('abc'), n),
          'date1': gen(pd.to_datetime('2010-01-01'),pd.to_datetime('2010-03-01'), n).floor('d'),
          'date2':gen(pd.to_datetime('2010-03-01'),pd.to_datetime('2010-06-01'), n).floor('d')
          )

          m = 5
          test2 = pd.DataFrame(
          'ID': np.random.choice(list('abc'), m),
          'amount':np.random.randint(10, size=m),
          'date':gen(pd.to_datetime('2010-01-01'), pd.to_datetime('2010-06-01'), m).floor('d')
          )



          print (test1)
          ID date1 date2
          0 c 2010-01-15 2010-05-22
          1 b 2010-02-08 2010-04-16
          2 c 2010-01-24 2010-04-12
          3 c 2010-02-01 2010-04-09
          4 a 2010-01-19 2010-05-20
          5 c 2010-01-27 2010-05-24
          6 c 2010-02-23 2010-03-15
          7 b 2010-01-31 2010-05-09
          8 c 2010-02-23 2010-03-29
          9 b 2010-01-08 2010-03-07

          print (test2)
          ID amount date
          0 a 4 2010-05-15
          1 b 6 2010-03-26
          2 a 1 2010-01-07
          3 b 5 2010-02-07
          4 a 6 2010-04-13

          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          print (df)
          ID date1 date2 amount date
          6 b 2010-02-08 2010-04-16 6.0 2010-03-26
          8 b 2010-01-31 2010-05-09 6.0 2010-03-26
          9 b 2010-01-31 2010-05-09 5.0 2010-02-07
          11 b 2010-01-08 2010-03-07 5.0 2010-02-07
          12 a 2010-01-19 2010-05-20 4.0 2010-05-15
          14 a 2010-01-19 2010-05-20 6.0 2010-04-13



          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])
          print (test)
          ID date1 date2 amount
          0 c 2010-01-15 2010-05-22 NaN
          1 b 2010-02-08 2010-04-16 6.0
          2 c 2010-01-24 2010-04-12 NaN
          3 c 2010-02-01 2010-04-09 NaN
          4 a 2010-01-19 2010-05-20 10.0
          5 c 2010-01-27 2010-05-24 NaN
          6 c 2010-02-23 2010-03-15 NaN
          7 b 2010-01-31 2010-05-09 11.0
          8 c 2010-02-23 2010-03-29 NaN
          9 b 2010-01-08 2010-03-07 5.0





          share|improve this answer

























          • What if an ID shows up more than once in test2? Will the amount be summed when you outer join?

            – jchaykow
            Nov 13 '18 at 7:17











          • @jchaykow - aded sample data for check it

            – jezrael
            Nov 13 '18 at 8:46










          Your Answer






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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          Use:



          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])


          Sample:



          #https://stackoverflow.com/q/21494489
          np.random.seed(123)

          #https://stackoverflow.com/a/50559321/2901002
          def gen(start, end, n):
          start_u = start.value//10**9
          end_u = end.value//10**9

          return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')

          n = 10
          test1 = pd.DataFrame(
          'ID':np.random.choice(list('abc'), n),
          'date1': gen(pd.to_datetime('2010-01-01'),pd.to_datetime('2010-03-01'), n).floor('d'),
          'date2':gen(pd.to_datetime('2010-03-01'),pd.to_datetime('2010-06-01'), n).floor('d')
          )

          m = 5
          test2 = pd.DataFrame(
          'ID': np.random.choice(list('abc'), m),
          'amount':np.random.randint(10, size=m),
          'date':gen(pd.to_datetime('2010-01-01'), pd.to_datetime('2010-06-01'), m).floor('d')
          )



          print (test1)
          ID date1 date2
          0 c 2010-01-15 2010-05-22
          1 b 2010-02-08 2010-04-16
          2 c 2010-01-24 2010-04-12
          3 c 2010-02-01 2010-04-09
          4 a 2010-01-19 2010-05-20
          5 c 2010-01-27 2010-05-24
          6 c 2010-02-23 2010-03-15
          7 b 2010-01-31 2010-05-09
          8 c 2010-02-23 2010-03-29
          9 b 2010-01-08 2010-03-07

          print (test2)
          ID amount date
          0 a 4 2010-05-15
          1 b 6 2010-03-26
          2 a 1 2010-01-07
          3 b 5 2010-02-07
          4 a 6 2010-04-13

          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          print (df)
          ID date1 date2 amount date
          6 b 2010-02-08 2010-04-16 6.0 2010-03-26
          8 b 2010-01-31 2010-05-09 6.0 2010-03-26
          9 b 2010-01-31 2010-05-09 5.0 2010-02-07
          11 b 2010-01-08 2010-03-07 5.0 2010-02-07
          12 a 2010-01-19 2010-05-20 4.0 2010-05-15
          14 a 2010-01-19 2010-05-20 6.0 2010-04-13



          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])
          print (test)
          ID date1 date2 amount
          0 c 2010-01-15 2010-05-22 NaN
          1 b 2010-02-08 2010-04-16 6.0
          2 c 2010-01-24 2010-04-12 NaN
          3 c 2010-02-01 2010-04-09 NaN
          4 a 2010-01-19 2010-05-20 10.0
          5 c 2010-01-27 2010-05-24 NaN
          6 c 2010-02-23 2010-03-15 NaN
          7 b 2010-01-31 2010-05-09 11.0
          8 c 2010-02-23 2010-03-29 NaN
          9 b 2010-01-08 2010-03-07 5.0





          share|improve this answer

























          • What if an ID shows up more than once in test2? Will the amount be summed when you outer join?

            – jchaykow
            Nov 13 '18 at 7:17











          • @jchaykow - aded sample data for check it

            – jezrael
            Nov 13 '18 at 8:46















          3














          Use:



          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])


          Sample:



          #https://stackoverflow.com/q/21494489
          np.random.seed(123)

          #https://stackoverflow.com/a/50559321/2901002
          def gen(start, end, n):
          start_u = start.value//10**9
          end_u = end.value//10**9

          return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')

          n = 10
          test1 = pd.DataFrame(
          'ID':np.random.choice(list('abc'), n),
          'date1': gen(pd.to_datetime('2010-01-01'),pd.to_datetime('2010-03-01'), n).floor('d'),
          'date2':gen(pd.to_datetime('2010-03-01'),pd.to_datetime('2010-06-01'), n).floor('d')
          )

          m = 5
          test2 = pd.DataFrame(
          'ID': np.random.choice(list('abc'), m),
          'amount':np.random.randint(10, size=m),
          'date':gen(pd.to_datetime('2010-01-01'), pd.to_datetime('2010-06-01'), m).floor('d')
          )



          print (test1)
          ID date1 date2
          0 c 2010-01-15 2010-05-22
          1 b 2010-02-08 2010-04-16
          2 c 2010-01-24 2010-04-12
          3 c 2010-02-01 2010-04-09
          4 a 2010-01-19 2010-05-20
          5 c 2010-01-27 2010-05-24
          6 c 2010-02-23 2010-03-15
          7 b 2010-01-31 2010-05-09
          8 c 2010-02-23 2010-03-29
          9 b 2010-01-08 2010-03-07

          print (test2)
          ID amount date
          0 a 4 2010-05-15
          1 b 6 2010-03-26
          2 a 1 2010-01-07
          3 b 5 2010-02-07
          4 a 6 2010-04-13

          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          print (df)
          ID date1 date2 amount date
          6 b 2010-02-08 2010-04-16 6.0 2010-03-26
          8 b 2010-01-31 2010-05-09 6.0 2010-03-26
          9 b 2010-01-31 2010-05-09 5.0 2010-02-07
          11 b 2010-01-08 2010-03-07 5.0 2010-02-07
          12 a 2010-01-19 2010-05-20 4.0 2010-05-15
          14 a 2010-01-19 2010-05-20 6.0 2010-04-13



          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])
          print (test)
          ID date1 date2 amount
          0 c 2010-01-15 2010-05-22 NaN
          1 b 2010-02-08 2010-04-16 6.0
          2 c 2010-01-24 2010-04-12 NaN
          3 c 2010-02-01 2010-04-09 NaN
          4 a 2010-01-19 2010-05-20 10.0
          5 c 2010-01-27 2010-05-24 NaN
          6 c 2010-02-23 2010-03-15 NaN
          7 b 2010-01-31 2010-05-09 11.0
          8 c 2010-02-23 2010-03-29 NaN
          9 b 2010-01-08 2010-03-07 5.0





          share|improve this answer

























          • What if an ID shows up more than once in test2? Will the amount be summed when you outer join?

            – jchaykow
            Nov 13 '18 at 7:17











          • @jchaykow - aded sample data for check it

            – jezrael
            Nov 13 '18 at 8:46













          3












          3








          3







          Use:



          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])


          Sample:



          #https://stackoverflow.com/q/21494489
          np.random.seed(123)

          #https://stackoverflow.com/a/50559321/2901002
          def gen(start, end, n):
          start_u = start.value//10**9
          end_u = end.value//10**9

          return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')

          n = 10
          test1 = pd.DataFrame(
          'ID':np.random.choice(list('abc'), n),
          'date1': gen(pd.to_datetime('2010-01-01'),pd.to_datetime('2010-03-01'), n).floor('d'),
          'date2':gen(pd.to_datetime('2010-03-01'),pd.to_datetime('2010-06-01'), n).floor('d')
          )

          m = 5
          test2 = pd.DataFrame(
          'ID': np.random.choice(list('abc'), m),
          'amount':np.random.randint(10, size=m),
          'date':gen(pd.to_datetime('2010-01-01'), pd.to_datetime('2010-06-01'), m).floor('d')
          )



          print (test1)
          ID date1 date2
          0 c 2010-01-15 2010-05-22
          1 b 2010-02-08 2010-04-16
          2 c 2010-01-24 2010-04-12
          3 c 2010-02-01 2010-04-09
          4 a 2010-01-19 2010-05-20
          5 c 2010-01-27 2010-05-24
          6 c 2010-02-23 2010-03-15
          7 b 2010-01-31 2010-05-09
          8 c 2010-02-23 2010-03-29
          9 b 2010-01-08 2010-03-07

          print (test2)
          ID amount date
          0 a 4 2010-05-15
          1 b 6 2010-03-26
          2 a 1 2010-01-07
          3 b 5 2010-02-07
          4 a 6 2010-04-13

          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          print (df)
          ID date1 date2 amount date
          6 b 2010-02-08 2010-04-16 6.0 2010-03-26
          8 b 2010-01-31 2010-05-09 6.0 2010-03-26
          9 b 2010-01-31 2010-05-09 5.0 2010-02-07
          11 b 2010-01-08 2010-03-07 5.0 2010-02-07
          12 a 2010-01-19 2010-05-20 4.0 2010-05-15
          14 a 2010-01-19 2010-05-20 6.0 2010-04-13



          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])
          print (test)
          ID date1 date2 amount
          0 c 2010-01-15 2010-05-22 NaN
          1 b 2010-02-08 2010-04-16 6.0
          2 c 2010-01-24 2010-04-12 NaN
          3 c 2010-02-01 2010-04-09 NaN
          4 a 2010-01-19 2010-05-20 10.0
          5 c 2010-01-27 2010-05-24 NaN
          6 c 2010-02-23 2010-03-15 NaN
          7 b 2010-01-31 2010-05-09 11.0
          8 c 2010-02-23 2010-03-29 NaN
          9 b 2010-01-08 2010-03-07 5.0





          share|improve this answer















          Use:



          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])


          Sample:



          #https://stackoverflow.com/q/21494489
          np.random.seed(123)

          #https://stackoverflow.com/a/50559321/2901002
          def gen(start, end, n):
          start_u = start.value//10**9
          end_u = end.value//10**9

          return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')

          n = 10
          test1 = pd.DataFrame(
          'ID':np.random.choice(list('abc'), n),
          'date1': gen(pd.to_datetime('2010-01-01'),pd.to_datetime('2010-03-01'), n).floor('d'),
          'date2':gen(pd.to_datetime('2010-03-01'),pd.to_datetime('2010-06-01'), n).floor('d')
          )

          m = 5
          test2 = pd.DataFrame(
          'ID': np.random.choice(list('abc'), m),
          'amount':np.random.randint(10, size=m),
          'date':gen(pd.to_datetime('2010-01-01'), pd.to_datetime('2010-06-01'), m).floor('d')
          )



          print (test1)
          ID date1 date2
          0 c 2010-01-15 2010-05-22
          1 b 2010-02-08 2010-04-16
          2 c 2010-01-24 2010-04-12
          3 c 2010-02-01 2010-04-09
          4 a 2010-01-19 2010-05-20
          5 c 2010-01-27 2010-05-24
          6 c 2010-02-23 2010-03-15
          7 b 2010-01-31 2010-05-09
          8 c 2010-02-23 2010-03-29
          9 b 2010-01-08 2010-03-07

          print (test2)
          ID amount date
          0 a 4 2010-05-15
          1 b 6 2010-03-26
          2 a 1 2010-01-07
          3 b 5 2010-02-07
          4 a 6 2010-04-13

          #outer join both df by ID columns
          df = test1.merge(test2, on='ID', how='outer')
          #filter by range
          df = df[(df.date > df.date1) & (df.date < df.date2)]
          print (df)
          ID date1 date2 amount date
          6 b 2010-02-08 2010-04-16 6.0 2010-03-26
          8 b 2010-01-31 2010-05-09 6.0 2010-03-26
          9 b 2010-01-31 2010-05-09 5.0 2010-02-07
          11 b 2010-01-08 2010-03-07 5.0 2010-02-07
          12 a 2010-01-19 2010-05-20 4.0 2010-05-15
          14 a 2010-01-19 2010-05-20 6.0 2010-04-13



          #thank you @Abhi for alternative
          #df = df[df.date.between(df.date1, df.date2, inclusive=False)]
          #aggregate sum
          s = df.groupby(['ID','date1','date2'])['amount'].sum()
          #add new column to test1
          test = test1.join(s, on=['ID','date1','date2'])
          print (test)
          ID date1 date2 amount
          0 c 2010-01-15 2010-05-22 NaN
          1 b 2010-02-08 2010-04-16 6.0
          2 c 2010-01-24 2010-04-12 NaN
          3 c 2010-02-01 2010-04-09 NaN
          4 a 2010-01-19 2010-05-20 10.0
          5 c 2010-01-27 2010-05-24 NaN
          6 c 2010-02-23 2010-03-15 NaN
          7 b 2010-01-31 2010-05-09 11.0
          8 c 2010-02-23 2010-03-29 NaN
          9 b 2010-01-08 2010-03-07 5.0






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 13 '18 at 8:46

























          answered Nov 13 '18 at 6:39









          jezraeljezrael

          331k24273351




          331k24273351












          • What if an ID shows up more than once in test2? Will the amount be summed when you outer join?

            – jchaykow
            Nov 13 '18 at 7:17











          • @jchaykow - aded sample data for check it

            – jezrael
            Nov 13 '18 at 8:46

















          • What if an ID shows up more than once in test2? Will the amount be summed when you outer join?

            – jchaykow
            Nov 13 '18 at 7:17











          • @jchaykow - aded sample data for check it

            – jezrael
            Nov 13 '18 at 8:46
















          What if an ID shows up more than once in test2? Will the amount be summed when you outer join?

          – jchaykow
          Nov 13 '18 at 7:17





          What if an ID shows up more than once in test2? Will the amount be summed when you outer join?

          – jchaykow
          Nov 13 '18 at 7:17













          @jchaykow - aded sample data for check it

          – jezrael
          Nov 13 '18 at 8:46





          @jchaykow - aded sample data for check it

          – jezrael
          Nov 13 '18 at 8:46

















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