Python column datetime subtract datetime












1















understand that there is many similar question out there on subtraction for pandas column, but my scenario is rather unique



I have 2 sets of data which both includes datetime. My program requires me to match this 2 date time therefore I used merge_asof to match to the closest time. After matching the data, I will need to find the time difference between this 2 set of data.



EndTime,Datetime
3/10/2010 0:00:33, 3/10/2010 0:00:26
3/10/2010 0:01:15,
3/10/2010 0:01:30,
3/10/2010 0:02:09, 3/10/2010 0:01:36
3/10/2010 0:02:50,
3/10/2010 0:05:09,
3/10/2010 0:06:00, 3/10/2010 0:05:48


The UNIQUE part will be that I would like to use Datetime subtract 1row before of endtime, (e.g. last row 3/10/2010 0:05:48 - 2nd last row 3/10/2010 0:05:09 = 39seconds)



Expected outcome would be:



EndTime,Datetime,SecondsDiff
3/10/2010 0:00:33, 3/10/2010 0:00:26, *not sure how to compute this but not important for 1st row of data*
3/10/2010 0:01:15,,
3/10/2010 0:01:30,,
3/10/2010 0:02:09, 3/10/2010 0:01:36,6
3/10/2010 0:02:50,,
3/10/2010 0:05:09,,
3/10/2010 0:06:00, 3/10/2010 0:05:48,39


I've tried some methods that caused some error, please advice!



As when I matched data, it cause duplicate so I uses



dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = ' '


to give empty space that doesnt cause duplicate but it give me "Error parsing datetime string " " at position 1"



-



I've also tried indicating the duplicates as 0 and converting the datetime type but causes the



error ufunc subtract cannot use operands with types dtype('O') anddtype('<M8[ns]')


, therefore I tried the below method but resulting to my datetime column becoming only date and the seconds difference was totally off



dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = 0
dm['EndTime'] = dm['EndTime'].values.astype('datetime64')
dm['Datetime'] = dm['Datetime'].values.astype('datetime64')
dm['Seconds'] = (dm.Datetime -
dm.EndTime.shift(-1)).astype('timedelta64[s]')









share|improve this question





























    1















    understand that there is many similar question out there on subtraction for pandas column, but my scenario is rather unique



    I have 2 sets of data which both includes datetime. My program requires me to match this 2 date time therefore I used merge_asof to match to the closest time. After matching the data, I will need to find the time difference between this 2 set of data.



    EndTime,Datetime
    3/10/2010 0:00:33, 3/10/2010 0:00:26
    3/10/2010 0:01:15,
    3/10/2010 0:01:30,
    3/10/2010 0:02:09, 3/10/2010 0:01:36
    3/10/2010 0:02:50,
    3/10/2010 0:05:09,
    3/10/2010 0:06:00, 3/10/2010 0:05:48


    The UNIQUE part will be that I would like to use Datetime subtract 1row before of endtime, (e.g. last row 3/10/2010 0:05:48 - 2nd last row 3/10/2010 0:05:09 = 39seconds)



    Expected outcome would be:



    EndTime,Datetime,SecondsDiff
    3/10/2010 0:00:33, 3/10/2010 0:00:26, *not sure how to compute this but not important for 1st row of data*
    3/10/2010 0:01:15,,
    3/10/2010 0:01:30,,
    3/10/2010 0:02:09, 3/10/2010 0:01:36,6
    3/10/2010 0:02:50,,
    3/10/2010 0:05:09,,
    3/10/2010 0:06:00, 3/10/2010 0:05:48,39


    I've tried some methods that caused some error, please advice!



    As when I matched data, it cause duplicate so I uses



    dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = ' '


    to give empty space that doesnt cause duplicate but it give me "Error parsing datetime string " " at position 1"



    -



    I've also tried indicating the duplicates as 0 and converting the datetime type but causes the



    error ufunc subtract cannot use operands with types dtype('O') anddtype('<M8[ns]')


    , therefore I tried the below method but resulting to my datetime column becoming only date and the seconds difference was totally off



    dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = 0
    dm['EndTime'] = dm['EndTime'].values.astype('datetime64')
    dm['Datetime'] = dm['Datetime'].values.astype('datetime64')
    dm['Seconds'] = (dm.Datetime -
    dm.EndTime.shift(-1)).astype('timedelta64[s]')









    share|improve this question



























      1












      1








      1








      understand that there is many similar question out there on subtraction for pandas column, but my scenario is rather unique



      I have 2 sets of data which both includes datetime. My program requires me to match this 2 date time therefore I used merge_asof to match to the closest time. After matching the data, I will need to find the time difference between this 2 set of data.



      EndTime,Datetime
      3/10/2010 0:00:33, 3/10/2010 0:00:26
      3/10/2010 0:01:15,
      3/10/2010 0:01:30,
      3/10/2010 0:02:09, 3/10/2010 0:01:36
      3/10/2010 0:02:50,
      3/10/2010 0:05:09,
      3/10/2010 0:06:00, 3/10/2010 0:05:48


      The UNIQUE part will be that I would like to use Datetime subtract 1row before of endtime, (e.g. last row 3/10/2010 0:05:48 - 2nd last row 3/10/2010 0:05:09 = 39seconds)



      Expected outcome would be:



      EndTime,Datetime,SecondsDiff
      3/10/2010 0:00:33, 3/10/2010 0:00:26, *not sure how to compute this but not important for 1st row of data*
      3/10/2010 0:01:15,,
      3/10/2010 0:01:30,,
      3/10/2010 0:02:09, 3/10/2010 0:01:36,6
      3/10/2010 0:02:50,,
      3/10/2010 0:05:09,,
      3/10/2010 0:06:00, 3/10/2010 0:05:48,39


      I've tried some methods that caused some error, please advice!



      As when I matched data, it cause duplicate so I uses



      dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = ' '


      to give empty space that doesnt cause duplicate but it give me "Error parsing datetime string " " at position 1"



      -



      I've also tried indicating the duplicates as 0 and converting the datetime type but causes the



      error ufunc subtract cannot use operands with types dtype('O') anddtype('<M8[ns]')


      , therefore I tried the below method but resulting to my datetime column becoming only date and the seconds difference was totally off



      dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = 0
      dm['EndTime'] = dm['EndTime'].values.astype('datetime64')
      dm['Datetime'] = dm['Datetime'].values.astype('datetime64')
      dm['Seconds'] = (dm.Datetime -
      dm.EndTime.shift(-1)).astype('timedelta64[s]')









      share|improve this question
















      understand that there is many similar question out there on subtraction for pandas column, but my scenario is rather unique



      I have 2 sets of data which both includes datetime. My program requires me to match this 2 date time therefore I used merge_asof to match to the closest time. After matching the data, I will need to find the time difference between this 2 set of data.



      EndTime,Datetime
      3/10/2010 0:00:33, 3/10/2010 0:00:26
      3/10/2010 0:01:15,
      3/10/2010 0:01:30,
      3/10/2010 0:02:09, 3/10/2010 0:01:36
      3/10/2010 0:02:50,
      3/10/2010 0:05:09,
      3/10/2010 0:06:00, 3/10/2010 0:05:48


      The UNIQUE part will be that I would like to use Datetime subtract 1row before of endtime, (e.g. last row 3/10/2010 0:05:48 - 2nd last row 3/10/2010 0:05:09 = 39seconds)



      Expected outcome would be:



      EndTime,Datetime,SecondsDiff
      3/10/2010 0:00:33, 3/10/2010 0:00:26, *not sure how to compute this but not important for 1st row of data*
      3/10/2010 0:01:15,,
      3/10/2010 0:01:30,,
      3/10/2010 0:02:09, 3/10/2010 0:01:36,6
      3/10/2010 0:02:50,,
      3/10/2010 0:05:09,,
      3/10/2010 0:06:00, 3/10/2010 0:05:48,39


      I've tried some methods that caused some error, please advice!



      As when I matched data, it cause duplicate so I uses



      dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = ' '


      to give empty space that doesnt cause duplicate but it give me "Error parsing datetime string " " at position 1"



      -



      I've also tried indicating the duplicates as 0 and converting the datetime type but causes the



      error ufunc subtract cannot use operands with types dtype('O') anddtype('<M8[ns]')


      , therefore I tried the below method but resulting to my datetime column becoming only date and the seconds difference was totally off



      dm.loc[(dm['Datetime'].notnull())&(dm.duplicated('Datetime')==True),'Datetime'] = 0
      dm['EndTime'] = dm['EndTime'].values.astype('datetime64')
      dm['Datetime'] = dm['Datetime'].values.astype('datetime64')
      dm['Seconds'] = (dm.Datetime -
      dm.EndTime.shift(-1)).astype('timedelta64[s]')






      python pandas datetime






      share|improve this question















      share|improve this question













      share|improve this question




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      edited Nov 23 '18 at 2:31









      Abhi

      2,540421




      2,540421










      asked Nov 23 '18 at 2:03









      ThanksForHelpingThanksForHelping

      647




      647
























          1 Answer
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          active

          oldest

          votes


















          0














          IIUC you can use :



          print (df)

          EndTime Datetime
          0 3/10/2010 0:00:33 3/10/2010 0:00:26
          1 3/10/2010 0:01:15
          2 3/10/2010 0:01:30 NaN
          3 3/10/2010 0:02:09 3/10/2010 0:01:36
          4 3/10/2010 0:02:50 NaN
          5 3/10/2010 0:05:09 NaN
          6 3/10/2010 0:06:00 3/10/2010 0:05:48




          df.Datetime = pd.to_datetime(df.Datetime, errors='coerce')
          df.EndTime = pd.to_datetime(df.EndTime, errors='coerce')

          df['SecondsDiff'] = df.Datetime.sub(df.EndTime.shift()).dt.seconds

          print (df)

          EndTime Datetime SecondsDiff
          0 2010-03-10 00:00:33 2010-03-10 00:00:26 NaN
          1 2010-03-10 00:01:15 NaT NaN
          2 2010-03-10 00:01:30 NaT NaN
          3 2010-03-10 00:02:09 2010-03-10 00:01:36 6.0
          4 2010-03-10 00:02:50 NaT NaN
          5 2010-03-10 00:05:09 NaT NaN
          6 2010-03-10 00:06:00 2010-03-10 00:05:48 39.0





          share|improve this answer























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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes









            0














            IIUC you can use :



            print (df)

            EndTime Datetime
            0 3/10/2010 0:00:33 3/10/2010 0:00:26
            1 3/10/2010 0:01:15
            2 3/10/2010 0:01:30 NaN
            3 3/10/2010 0:02:09 3/10/2010 0:01:36
            4 3/10/2010 0:02:50 NaN
            5 3/10/2010 0:05:09 NaN
            6 3/10/2010 0:06:00 3/10/2010 0:05:48




            df.Datetime = pd.to_datetime(df.Datetime, errors='coerce')
            df.EndTime = pd.to_datetime(df.EndTime, errors='coerce')

            df['SecondsDiff'] = df.Datetime.sub(df.EndTime.shift()).dt.seconds

            print (df)

            EndTime Datetime SecondsDiff
            0 2010-03-10 00:00:33 2010-03-10 00:00:26 NaN
            1 2010-03-10 00:01:15 NaT NaN
            2 2010-03-10 00:01:30 NaT NaN
            3 2010-03-10 00:02:09 2010-03-10 00:01:36 6.0
            4 2010-03-10 00:02:50 NaT NaN
            5 2010-03-10 00:05:09 NaT NaN
            6 2010-03-10 00:06:00 2010-03-10 00:05:48 39.0





            share|improve this answer




























              0














              IIUC you can use :



              print (df)

              EndTime Datetime
              0 3/10/2010 0:00:33 3/10/2010 0:00:26
              1 3/10/2010 0:01:15
              2 3/10/2010 0:01:30 NaN
              3 3/10/2010 0:02:09 3/10/2010 0:01:36
              4 3/10/2010 0:02:50 NaN
              5 3/10/2010 0:05:09 NaN
              6 3/10/2010 0:06:00 3/10/2010 0:05:48




              df.Datetime = pd.to_datetime(df.Datetime, errors='coerce')
              df.EndTime = pd.to_datetime(df.EndTime, errors='coerce')

              df['SecondsDiff'] = df.Datetime.sub(df.EndTime.shift()).dt.seconds

              print (df)

              EndTime Datetime SecondsDiff
              0 2010-03-10 00:00:33 2010-03-10 00:00:26 NaN
              1 2010-03-10 00:01:15 NaT NaN
              2 2010-03-10 00:01:30 NaT NaN
              3 2010-03-10 00:02:09 2010-03-10 00:01:36 6.0
              4 2010-03-10 00:02:50 NaT NaN
              5 2010-03-10 00:05:09 NaT NaN
              6 2010-03-10 00:06:00 2010-03-10 00:05:48 39.0





              share|improve this answer


























                0












                0








                0







                IIUC you can use :



                print (df)

                EndTime Datetime
                0 3/10/2010 0:00:33 3/10/2010 0:00:26
                1 3/10/2010 0:01:15
                2 3/10/2010 0:01:30 NaN
                3 3/10/2010 0:02:09 3/10/2010 0:01:36
                4 3/10/2010 0:02:50 NaN
                5 3/10/2010 0:05:09 NaN
                6 3/10/2010 0:06:00 3/10/2010 0:05:48




                df.Datetime = pd.to_datetime(df.Datetime, errors='coerce')
                df.EndTime = pd.to_datetime(df.EndTime, errors='coerce')

                df['SecondsDiff'] = df.Datetime.sub(df.EndTime.shift()).dt.seconds

                print (df)

                EndTime Datetime SecondsDiff
                0 2010-03-10 00:00:33 2010-03-10 00:00:26 NaN
                1 2010-03-10 00:01:15 NaT NaN
                2 2010-03-10 00:01:30 NaT NaN
                3 2010-03-10 00:02:09 2010-03-10 00:01:36 6.0
                4 2010-03-10 00:02:50 NaT NaN
                5 2010-03-10 00:05:09 NaT NaN
                6 2010-03-10 00:06:00 2010-03-10 00:05:48 39.0





                share|improve this answer













                IIUC you can use :



                print (df)

                EndTime Datetime
                0 3/10/2010 0:00:33 3/10/2010 0:00:26
                1 3/10/2010 0:01:15
                2 3/10/2010 0:01:30 NaN
                3 3/10/2010 0:02:09 3/10/2010 0:01:36
                4 3/10/2010 0:02:50 NaN
                5 3/10/2010 0:05:09 NaN
                6 3/10/2010 0:06:00 3/10/2010 0:05:48




                df.Datetime = pd.to_datetime(df.Datetime, errors='coerce')
                df.EndTime = pd.to_datetime(df.EndTime, errors='coerce')

                df['SecondsDiff'] = df.Datetime.sub(df.EndTime.shift()).dt.seconds

                print (df)

                EndTime Datetime SecondsDiff
                0 2010-03-10 00:00:33 2010-03-10 00:00:26 NaN
                1 2010-03-10 00:01:15 NaT NaN
                2 2010-03-10 00:01:30 NaT NaN
                3 2010-03-10 00:02:09 2010-03-10 00:01:36 6.0
                4 2010-03-10 00:02:50 NaT NaN
                5 2010-03-10 00:05:09 NaT NaN
                6 2010-03-10 00:06:00 2010-03-10 00:05:48 39.0






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 23 '18 at 2:27









                AbhiAbhi

                2,540421




                2,540421
































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