Convert dataframe to quarter











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I have a dataframe (df). The Date variable is a string. I want to convert it to date and reformat it to the date at the end of the quarter. A sample is below:



df:
Date
0 201601
1 201602
2 201603
3 201604


201601 is the first quarter of 2016 and 201604 is the fourth quarter of 2016. The desired outcome is:



df:
Date
0 2016-03-31
1 2016-06-30
2 2016-09-30
3 2016-12-31


This is what I tried, but it doesn't work.



df['date'] = pd.to_datetime(df.Date, format = '%Y%q')


Thanks!










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    up vote
    1
    down vote

    favorite












    I have a dataframe (df). The Date variable is a string. I want to convert it to date and reformat it to the date at the end of the quarter. A sample is below:



    df:
    Date
    0 201601
    1 201602
    2 201603
    3 201604


    201601 is the first quarter of 2016 and 201604 is the fourth quarter of 2016. The desired outcome is:



    df:
    Date
    0 2016-03-31
    1 2016-06-30
    2 2016-09-30
    3 2016-12-31


    This is what I tried, but it doesn't work.



    df['date'] = pd.to_datetime(df.Date, format = '%Y%q')


    Thanks!










    share|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have a dataframe (df). The Date variable is a string. I want to convert it to date and reformat it to the date at the end of the quarter. A sample is below:



      df:
      Date
      0 201601
      1 201602
      2 201603
      3 201604


      201601 is the first quarter of 2016 and 201604 is the fourth quarter of 2016. The desired outcome is:



      df:
      Date
      0 2016-03-31
      1 2016-06-30
      2 2016-09-30
      3 2016-12-31


      This is what I tried, but it doesn't work.



      df['date'] = pd.to_datetime(df.Date, format = '%Y%q')


      Thanks!










      share|improve this question













      I have a dataframe (df). The Date variable is a string. I want to convert it to date and reformat it to the date at the end of the quarter. A sample is below:



      df:
      Date
      0 201601
      1 201602
      2 201603
      3 201604


      201601 is the first quarter of 2016 and 201604 is the fourth quarter of 2016. The desired outcome is:



      df:
      Date
      0 2016-03-31
      1 2016-06-30
      2 2016-09-30
      3 2016-12-31


      This is what I tried, but it doesn't work.



      df['date'] = pd.to_datetime(df.Date, format = '%Y%q')


      Thanks!







      python






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      asked Nov 19 at 14:20









      Roger

      5911




      5911
























          2 Answers
          2






          active

          oldest

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          up vote
          1
          down vote













          You could define a function to compute the date and then apply that function.



          Using the monthrange function from the calendar module (https://docs.python.org/3.7/library/calendar.html#calendar.monthrange), your function might look something like:



          from calendar import monthrange
          from datetime import datetime

          def end_quarter(quarter):
          year = int(quarter[:4])
          month = int(quarter[-2:]) * 3
          day = monthrange(year, month)[1]
          return datetime(year, month, day).date()


          and you could then apply it using:



          df["Date"] = df["Date"].apply(end_quarter)





          share|improve this answer






























            up vote
            0
            down vote













            One way is to keep treating it as a string, and simply do an if / else statement, i.e.



            res = 
            for i in df['Date']:
            v1 = i[-2:]
            if v1 == '01':
            res.append(i[:4] + '-03-31')
            elif v1 == '02':
            res.append(i[:4] + '-06-30')
            elif v1 == '03':
            res.append(i[4:] + '-09-30')
            else:
            res.append(i[4:] + '-12-31')

            #>>> res
            #['2016-03-31', '2016-06-30', '03-09-30', '04-12-31']





            share|improve this answer





















              Your Answer






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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote













              You could define a function to compute the date and then apply that function.



              Using the monthrange function from the calendar module (https://docs.python.org/3.7/library/calendar.html#calendar.monthrange), your function might look something like:



              from calendar import monthrange
              from datetime import datetime

              def end_quarter(quarter):
              year = int(quarter[:4])
              month = int(quarter[-2:]) * 3
              day = monthrange(year, month)[1]
              return datetime(year, month, day).date()


              and you could then apply it using:



              df["Date"] = df["Date"].apply(end_quarter)





              share|improve this answer



























                up vote
                1
                down vote













                You could define a function to compute the date and then apply that function.



                Using the monthrange function from the calendar module (https://docs.python.org/3.7/library/calendar.html#calendar.monthrange), your function might look something like:



                from calendar import monthrange
                from datetime import datetime

                def end_quarter(quarter):
                year = int(quarter[:4])
                month = int(quarter[-2:]) * 3
                day = monthrange(year, month)[1]
                return datetime(year, month, day).date()


                and you could then apply it using:



                df["Date"] = df["Date"].apply(end_quarter)





                share|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  You could define a function to compute the date and then apply that function.



                  Using the monthrange function from the calendar module (https://docs.python.org/3.7/library/calendar.html#calendar.monthrange), your function might look something like:



                  from calendar import monthrange
                  from datetime import datetime

                  def end_quarter(quarter):
                  year = int(quarter[:4])
                  month = int(quarter[-2:]) * 3
                  day = monthrange(year, month)[1]
                  return datetime(year, month, day).date()


                  and you could then apply it using:



                  df["Date"] = df["Date"].apply(end_quarter)





                  share|improve this answer














                  You could define a function to compute the date and then apply that function.



                  Using the monthrange function from the calendar module (https://docs.python.org/3.7/library/calendar.html#calendar.monthrange), your function might look something like:



                  from calendar import monthrange
                  from datetime import datetime

                  def end_quarter(quarter):
                  year = int(quarter[:4])
                  month = int(quarter[-2:]) * 3
                  day = monthrange(year, month)[1]
                  return datetime(year, month, day).date()


                  and you could then apply it using:



                  df["Date"] = df["Date"].apply(end_quarter)






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 20 at 10:26

























                  answered Nov 19 at 14:48









                  Owen

                  3,1301914




                  3,1301914
























                      up vote
                      0
                      down vote













                      One way is to keep treating it as a string, and simply do an if / else statement, i.e.



                      res = 
                      for i in df['Date']:
                      v1 = i[-2:]
                      if v1 == '01':
                      res.append(i[:4] + '-03-31')
                      elif v1 == '02':
                      res.append(i[:4] + '-06-30')
                      elif v1 == '03':
                      res.append(i[4:] + '-09-30')
                      else:
                      res.append(i[4:] + '-12-31')

                      #>>> res
                      #['2016-03-31', '2016-06-30', '03-09-30', '04-12-31']





                      share|improve this answer

























                        up vote
                        0
                        down vote













                        One way is to keep treating it as a string, and simply do an if / else statement, i.e.



                        res = 
                        for i in df['Date']:
                        v1 = i[-2:]
                        if v1 == '01':
                        res.append(i[:4] + '-03-31')
                        elif v1 == '02':
                        res.append(i[:4] + '-06-30')
                        elif v1 == '03':
                        res.append(i[4:] + '-09-30')
                        else:
                        res.append(i[4:] + '-12-31')

                        #>>> res
                        #['2016-03-31', '2016-06-30', '03-09-30', '04-12-31']





                        share|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          One way is to keep treating it as a string, and simply do an if / else statement, i.e.



                          res = 
                          for i in df['Date']:
                          v1 = i[-2:]
                          if v1 == '01':
                          res.append(i[:4] + '-03-31')
                          elif v1 == '02':
                          res.append(i[:4] + '-06-30')
                          elif v1 == '03':
                          res.append(i[4:] + '-09-30')
                          else:
                          res.append(i[4:] + '-12-31')

                          #>>> res
                          #['2016-03-31', '2016-06-30', '03-09-30', '04-12-31']





                          share|improve this answer












                          One way is to keep treating it as a string, and simply do an if / else statement, i.e.



                          res = 
                          for i in df['Date']:
                          v1 = i[-2:]
                          if v1 == '01':
                          res.append(i[:4] + '-03-31')
                          elif v1 == '02':
                          res.append(i[:4] + '-06-30')
                          elif v1 == '03':
                          res.append(i[4:] + '-09-30')
                          else:
                          res.append(i[4:] + '-12-31')

                          #>>> res
                          #['2016-03-31', '2016-06-30', '03-09-30', '04-12-31']






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 19 at 14:42









                          Sotos

                          26.9k51540




                          26.9k51540






























                               

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