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How to generate a unix time column from “year”,“date”,“month”,“time” columns, preferrably using pandas module for python


Does pandas iterrows have performance issues?How to get file creation & modification date/times in Python?How to get the current time in PythonHow can I make a time delay in Python?How to remove an element from a list by index?How to remove a key from a Python dictionary?Adding new column to existing DataFrame in Python pandasDelete column from pandas DataFrameSelect rows from a DataFrame based on values in a column in pandasGet list from pandas DataFrame column headersModify Pandas dataframe to list year month and date






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1















There is an excel file logging a set of data. Its columns are as below, where each column is seperated by comma.



SampleData
year,date,month,location,time,count
2019,20,Jan,Japan,22:33,1
2019,31,Jan,Japan,19:21,1
2019,1,Jan,Japan,8:00,1
2019,4,Jan,Japan,4:28,2
2019,13,Feb,Japan,6:19,1


From this data, I would like to create python pandas dataframe, which looks like below.



DataFrame
u_datetime,location,count
1547991180,Japan,1
1548930060,Japan,1
1546297200,Japan,1
1546543680,Japan,2
1550006340,Japan,1


One of the DataFrame methods can be useful for this operation, but it does not take date with one digit.



pandas.to_datetime(
DataFrame["year"].astype(str)
+ DataFrame["month"].astype(str)
+ DataFrame["date"].astype(str)
+ DataFrame["time"].astype(str),
format="%Y%b%d%-H%M"
)


Could anybody give me a hand?



Thank you.










share|improve this question






























    1















    There is an excel file logging a set of data. Its columns are as below, where each column is seperated by comma.



    SampleData
    year,date,month,location,time,count
    2019,20,Jan,Japan,22:33,1
    2019,31,Jan,Japan,19:21,1
    2019,1,Jan,Japan,8:00,1
    2019,4,Jan,Japan,4:28,2
    2019,13,Feb,Japan,6:19,1


    From this data, I would like to create python pandas dataframe, which looks like below.



    DataFrame
    u_datetime,location,count
    1547991180,Japan,1
    1548930060,Japan,1
    1546297200,Japan,1
    1546543680,Japan,2
    1550006340,Japan,1


    One of the DataFrame methods can be useful for this operation, but it does not take date with one digit.



    pandas.to_datetime(
    DataFrame["year"].astype(str)
    + DataFrame["month"].astype(str)
    + DataFrame["date"].astype(str)
    + DataFrame["time"].astype(str),
    format="%Y%b%d%-H%M"
    )


    Could anybody give me a hand?



    Thank you.










    share|improve this question


























      1












      1








      1








      There is an excel file logging a set of data. Its columns are as below, where each column is seperated by comma.



      SampleData
      year,date,month,location,time,count
      2019,20,Jan,Japan,22:33,1
      2019,31,Jan,Japan,19:21,1
      2019,1,Jan,Japan,8:00,1
      2019,4,Jan,Japan,4:28,2
      2019,13,Feb,Japan,6:19,1


      From this data, I would like to create python pandas dataframe, which looks like below.



      DataFrame
      u_datetime,location,count
      1547991180,Japan,1
      1548930060,Japan,1
      1546297200,Japan,1
      1546543680,Japan,2
      1550006340,Japan,1


      One of the DataFrame methods can be useful for this operation, but it does not take date with one digit.



      pandas.to_datetime(
      DataFrame["year"].astype(str)
      + DataFrame["month"].astype(str)
      + DataFrame["date"].astype(str)
      + DataFrame["time"].astype(str),
      format="%Y%b%d%-H%M"
      )


      Could anybody give me a hand?



      Thank you.










      share|improve this question
















      There is an excel file logging a set of data. Its columns are as below, where each column is seperated by comma.



      SampleData
      year,date,month,location,time,count
      2019,20,Jan,Japan,22:33,1
      2019,31,Jan,Japan,19:21,1
      2019,1,Jan,Japan,8:00,1
      2019,4,Jan,Japan,4:28,2
      2019,13,Feb,Japan,6:19,1


      From this data, I would like to create python pandas dataframe, which looks like below.



      DataFrame
      u_datetime,location,count
      1547991180,Japan,1
      1548930060,Japan,1
      1546297200,Japan,1
      1546543680,Japan,2
      1550006340,Japan,1


      One of the DataFrame methods can be useful for this operation, but it does not take date with one digit.



      pandas.to_datetime(
      DataFrame["year"].astype(str)
      + DataFrame["month"].astype(str)
      + DataFrame["date"].astype(str)
      + DataFrame["time"].astype(str),
      format="%Y%b%d%-H%M"
      )


      Could anybody give me a hand?



      Thank you.







      python pandas






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 26 at 7:00









      Erik Cederstrand

      4,0393 gold badges23 silver badges48 bronze badges




      4,0393 gold badges23 silver badges48 bronze badges










      asked Mar 26 at 6:31









      wisywisy

      61 bronze badge




      61 bronze badge






















          3 Answers
          3






          active

          oldest

          votes


















          1














          try this



          from datetime import datetime

          data['datetime'] = data[['year','date','month','time']].apply(lambda x: datetime.strptime(str(x['year'])+'-'+str(x['date'])+'-'+str(x['month'])+' '+str(x['time']), "%Y-%d-%b %H:%M").timestamp(), axis=1)

          data[['datetime','location','count']]


          Output



           datetime 
          0 1548003780.0
          1 1548942660.0
          2 1546309800.0
          3 1546556280.0
          4 1550018940.0

          location
          0 Japan
          1 Japan
          2 Japan
          3 Japan
          4 Japan

          count
          0 1
          1 1
          2 1
          3 2
          4 1





          share|improve this answer

























          • Yes, your solution working, but OP need pandas solution, not python. Also apply are loops under the hood, so better not use if exist vectorized solutions, here function pd.to_datetime. link

            – jezrael
            Mar 26 at 7:22



















          1














          In case you are working with csv file this can be done easily using parse_dates.



          dateparse = lambda x: pd.datetime.strptime(x, '%Y-%m-%d %H:%M:%S')
          df = pd.read_csv('/home/users/user/xxx.csv', parse_dates ='date_time':[0,1,2,4])
          df['u_datetime'] = df['date_time'].values.astype(np.int64) // 10 ** 9
          df_new = df[['u_datetime', 'location', 'count']]





          share|improve this answer






























            0














            You are close, need %Y%b%d%H:%M format and then convert to unix time by cast to int64 with integer division by 10**9:



            s = (DataFrame["year"].astype(str)+
            DataFrame["month"].astype(str)+
            DataFrame["date"].astype(str)+
            DataFrame["time"].astype(str))
            DataFrame['u_datetime'] = pd.to_datetime(s, format="%Y%b%d%H:%M").astype(np.int64) // 10**9

            DataFrame = DataFrame[['u_datetime','location','count']]
            print (DataFrame)
            u_datetime location count
            0 1548023580 Japan 1
            1 1548962460 Japan 1
            2 1546329600 Japan 1
            3 1546576080 Japan 2
            4 1550038740 Japan 1





            share|improve this answer

























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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              try this



              from datetime import datetime

              data['datetime'] = data[['year','date','month','time']].apply(lambda x: datetime.strptime(str(x['year'])+'-'+str(x['date'])+'-'+str(x['month'])+' '+str(x['time']), "%Y-%d-%b %H:%M").timestamp(), axis=1)

              data[['datetime','location','count']]


              Output



               datetime 
              0 1548003780.0
              1 1548942660.0
              2 1546309800.0
              3 1546556280.0
              4 1550018940.0

              location
              0 Japan
              1 Japan
              2 Japan
              3 Japan
              4 Japan

              count
              0 1
              1 1
              2 1
              3 2
              4 1





              share|improve this answer

























              • Yes, your solution working, but OP need pandas solution, not python. Also apply are loops under the hood, so better not use if exist vectorized solutions, here function pd.to_datetime. link

                – jezrael
                Mar 26 at 7:22
















              1














              try this



              from datetime import datetime

              data['datetime'] = data[['year','date','month','time']].apply(lambda x: datetime.strptime(str(x['year'])+'-'+str(x['date'])+'-'+str(x['month'])+' '+str(x['time']), "%Y-%d-%b %H:%M").timestamp(), axis=1)

              data[['datetime','location','count']]


              Output



               datetime 
              0 1548003780.0
              1 1548942660.0
              2 1546309800.0
              3 1546556280.0
              4 1550018940.0

              location
              0 Japan
              1 Japan
              2 Japan
              3 Japan
              4 Japan

              count
              0 1
              1 1
              2 1
              3 2
              4 1





              share|improve this answer

























              • Yes, your solution working, but OP need pandas solution, not python. Also apply are loops under the hood, so better not use if exist vectorized solutions, here function pd.to_datetime. link

                – jezrael
                Mar 26 at 7:22














              1












              1








              1







              try this



              from datetime import datetime

              data['datetime'] = data[['year','date','month','time']].apply(lambda x: datetime.strptime(str(x['year'])+'-'+str(x['date'])+'-'+str(x['month'])+' '+str(x['time']), "%Y-%d-%b %H:%M").timestamp(), axis=1)

              data[['datetime','location','count']]


              Output



               datetime 
              0 1548003780.0
              1 1548942660.0
              2 1546309800.0
              3 1546556280.0
              4 1550018940.0

              location
              0 Japan
              1 Japan
              2 Japan
              3 Japan
              4 Japan

              count
              0 1
              1 1
              2 1
              3 2
              4 1





              share|improve this answer















              try this



              from datetime import datetime

              data['datetime'] = data[['year','date','month','time']].apply(lambda x: datetime.strptime(str(x['year'])+'-'+str(x['date'])+'-'+str(x['month'])+' '+str(x['time']), "%Y-%d-%b %H:%M").timestamp(), axis=1)

              data[['datetime','location','count']]


              Output



               datetime 
              0 1548003780.0
              1 1548942660.0
              2 1546309800.0
              3 1546556280.0
              4 1550018940.0

              location
              0 Japan
              1 Japan
              2 Japan
              3 Japan
              4 Japan

              count
              0 1
              1 1
              2 1
              3 2
              4 1






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Mar 26 at 7:18

























              answered Mar 26 at 6:46









              iamklausiamklaus

              2,3761 gold badge6 silver badges18 bronze badges




              2,3761 gold badge6 silver badges18 bronze badges












              • Yes, your solution working, but OP need pandas solution, not python. Also apply are loops under the hood, so better not use if exist vectorized solutions, here function pd.to_datetime. link

                – jezrael
                Mar 26 at 7:22


















              • Yes, your solution working, but OP need pandas solution, not python. Also apply are loops under the hood, so better not use if exist vectorized solutions, here function pd.to_datetime. link

                – jezrael
                Mar 26 at 7:22

















              Yes, your solution working, but OP need pandas solution, not python. Also apply are loops under the hood, so better not use if exist vectorized solutions, here function pd.to_datetime. link

              – jezrael
              Mar 26 at 7:22






              Yes, your solution working, but OP need pandas solution, not python. Also apply are loops under the hood, so better not use if exist vectorized solutions, here function pd.to_datetime. link

              – jezrael
              Mar 26 at 7:22














              1














              In case you are working with csv file this can be done easily using parse_dates.



              dateparse = lambda x: pd.datetime.strptime(x, '%Y-%m-%d %H:%M:%S')
              df = pd.read_csv('/home/users/user/xxx.csv', parse_dates ='date_time':[0,1,2,4])
              df['u_datetime'] = df['date_time'].values.astype(np.int64) // 10 ** 9
              df_new = df[['u_datetime', 'location', 'count']]





              share|improve this answer



























                1














                In case you are working with csv file this can be done easily using parse_dates.



                dateparse = lambda x: pd.datetime.strptime(x, '%Y-%m-%d %H:%M:%S')
                df = pd.read_csv('/home/users/user/xxx.csv', parse_dates ='date_time':[0,1,2,4])
                df['u_datetime'] = df['date_time'].values.astype(np.int64) // 10 ** 9
                df_new = df[['u_datetime', 'location', 'count']]





                share|improve this answer

























                  1












                  1








                  1







                  In case you are working with csv file this can be done easily using parse_dates.



                  dateparse = lambda x: pd.datetime.strptime(x, '%Y-%m-%d %H:%M:%S')
                  df = pd.read_csv('/home/users/user/xxx.csv', parse_dates ='date_time':[0,1,2,4])
                  df['u_datetime'] = df['date_time'].values.astype(np.int64) // 10 ** 9
                  df_new = df[['u_datetime', 'location', 'count']]





                  share|improve this answer













                  In case you are working with csv file this can be done easily using parse_dates.



                  dateparse = lambda x: pd.datetime.strptime(x, '%Y-%m-%d %H:%M:%S')
                  df = pd.read_csv('/home/users/user/xxx.csv', parse_dates ='date_time':[0,1,2,4])
                  df['u_datetime'] = df['date_time'].values.astype(np.int64) // 10 ** 9
                  df_new = df[['u_datetime', 'location', 'count']]






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Mar 26 at 8:20









                  LoochieLoochie

                  1,0063 silver badges11 bronze badges




                  1,0063 silver badges11 bronze badges





















                      0














                      You are close, need %Y%b%d%H:%M format and then convert to unix time by cast to int64 with integer division by 10**9:



                      s = (DataFrame["year"].astype(str)+
                      DataFrame["month"].astype(str)+
                      DataFrame["date"].astype(str)+
                      DataFrame["time"].astype(str))
                      DataFrame['u_datetime'] = pd.to_datetime(s, format="%Y%b%d%H:%M").astype(np.int64) // 10**9

                      DataFrame = DataFrame[['u_datetime','location','count']]
                      print (DataFrame)
                      u_datetime location count
                      0 1548023580 Japan 1
                      1 1548962460 Japan 1
                      2 1546329600 Japan 1
                      3 1546576080 Japan 2
                      4 1550038740 Japan 1





                      share|improve this answer



























                        0














                        You are close, need %Y%b%d%H:%M format and then convert to unix time by cast to int64 with integer division by 10**9:



                        s = (DataFrame["year"].astype(str)+
                        DataFrame["month"].astype(str)+
                        DataFrame["date"].astype(str)+
                        DataFrame["time"].astype(str))
                        DataFrame['u_datetime'] = pd.to_datetime(s, format="%Y%b%d%H:%M").astype(np.int64) // 10**9

                        DataFrame = DataFrame[['u_datetime','location','count']]
                        print (DataFrame)
                        u_datetime location count
                        0 1548023580 Japan 1
                        1 1548962460 Japan 1
                        2 1546329600 Japan 1
                        3 1546576080 Japan 2
                        4 1550038740 Japan 1





                        share|improve this answer

























                          0












                          0








                          0







                          You are close, need %Y%b%d%H:%M format and then convert to unix time by cast to int64 with integer division by 10**9:



                          s = (DataFrame["year"].astype(str)+
                          DataFrame["month"].astype(str)+
                          DataFrame["date"].astype(str)+
                          DataFrame["time"].astype(str))
                          DataFrame['u_datetime'] = pd.to_datetime(s, format="%Y%b%d%H:%M").astype(np.int64) // 10**9

                          DataFrame = DataFrame[['u_datetime','location','count']]
                          print (DataFrame)
                          u_datetime location count
                          0 1548023580 Japan 1
                          1 1548962460 Japan 1
                          2 1546329600 Japan 1
                          3 1546576080 Japan 2
                          4 1550038740 Japan 1





                          share|improve this answer













                          You are close, need %Y%b%d%H:%M format and then convert to unix time by cast to int64 with integer division by 10**9:



                          s = (DataFrame["year"].astype(str)+
                          DataFrame["month"].astype(str)+
                          DataFrame["date"].astype(str)+
                          DataFrame["time"].astype(str))
                          DataFrame['u_datetime'] = pd.to_datetime(s, format="%Y%b%d%H:%M").astype(np.int64) // 10**9

                          DataFrame = DataFrame[['u_datetime','location','count']]
                          print (DataFrame)
                          u_datetime location count
                          0 1548023580 Japan 1
                          1 1548962460 Japan 1
                          2 1546329600 Japan 1
                          3 1546576080 Japan 2
                          4 1550038740 Japan 1






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Mar 26 at 6:50









                          jezraeljezrael

                          388k28 gold badges391 silver badges467 bronze badges




                          388k28 gold badges391 silver badges467 bronze badges



























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