How to calculate the time interval between two time strings

Question

I have two times  a start and a stop time  in the format of 10 33 26  HH MM SS    I need the difference between the two times   I ve been looking through documentation for Python and searching online and I would imagine it would have something to do with the datetime and or time modules   I can t get it to work properly and keep finding only how to do this when a date is involved   Ultimately  I need to calculate the averages of multiple time durations  I got the time differences to work and I m storing them in a list  I now need to calculate the average  I m using regular expressions to parse out the original times and then doing the differences     For the averaging  should I convert to seconds and then average

User · Answer

Try this  import datetime import time start time   datetime datetime now   time   strftime   H  M  S   time sleep 5  end time   datetime datetime now   time   strftime   H  M  S   total time  datetime datetime strptime end time   H  M  S   - datetime datetime strptime start time   H  M  S    print total time   OUTPUT    0 00 05

User · Answer

Here s a solution that supports finding the difference even if the end time is less than the start time  over midnight interval  such as 23 55 00-00 25 00  a half an hour duration       usr bin env python from datetime import datetime  time as datetime time  timedelta  def time diff start  end       if isinstance start  datetime time     convert to datetime         assert isinstance end  datetime time          start  end    datetime combine datetime min  t  for t in  start  end       if start  lt   end    e g   10 33 26-11 15 49         return end - start     else    end  lt  start e g   23 55 00-00 25 00         end    timedelta 1     day         assert end  gt  start         return end - start  for time range in   10 33 26-11 15 49    23 55 00-00 25 00        s  e    datetime strptime t    H  M  S   for t in time range split  -        print time diff s  e       assert time diff s  e     time diff s time    e time      Output  0 42 23 0 30 00   time diff   returns a timedelta object that you can pass  as a part of the sequence  to a mean   function directly e g       usr bin env python from datetime import timedelta  def mean data  start timedelta 0           Find arithmetic average         return sum data  start    len data   data    timedelta minutes 42  seconds 23     0 42 23         timedelta minutes 30     0 30 00 print repr mean data      - gt  datetime timedelta 0  2171  500000    days  seconds  microseconds   The mean   result is also timedelta   object that you can convert to seconds  td total seconds   method  since Python 2 7    hours  td   timedelta hours 1   Python 3    etc

User · Answer

import datetime as dt from dateutil relativedelta import relativedelta  start    09 35 23  end    10 23 00  start dt   dt datetime strptime start    H  M  S   end dt   dt datetime strptime end    H  M  S   timedelta obj   relativedelta start dt  end dt  print      timedelta obj years      timedelta obj months      timedelta obj days      timedelta obj hours      timedelta obj minutes      timedelta obj seconds      result   0 0 0 0 -47 -37

User · Answer

Concise if you are just interested in the time elapsed that is under 24 hours  You can format the output as needed in the return statement    import datetime def elapsed interval start end       elapsed   end - start     min secs divmod elapsed days   86400   elapsed seconds  60      hour  minutes   divmod min  60      return    2d   2d   2d     hour minutes secs   if   name         main         time start datetime datetime now           do your process         time end datetime datetime now       total time elapsed interval time start time end

User · Answer

Try this -- it s efficient for timing short-term events   If something takes more than an hour  then the final display probably will want some friendly formatting   import time start   time time    time sleep 10     or do something more productive  done   time time   elapsed   done - start print elapsed    The time difference is returned as the number of elapsed seconds

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import datetime          day   int input  quot day 1 2 3   31    quot        month   int input  quot Month 1 2 3    12    quot        year   int input  quot year 0 2020    quot        start date   datetime date year  month  day           day   int input  quot day 1 2 3   31    quot        month   int input  quot Month 1 2 3    12    quot        year   int input  quot year 0 2020    quot        end date   datetime date year  month  day           time difference   end date - start date     age   time difference days     print  quot Total days   quot    str age

User · Answer

Take a look at the datetime module and the timedelta objects   You should end up constructing a datetime object for the start and stop times  and when you subtract them  you get a timedelta

User · Answer

I like how this guy does it     https   amalgjose com 2015 02 19 python-code-for-calculating-the-difference-between-two-time-stamps  Not sure if it has some cons   But looks neat for me     from datetime import datetime from dateutil relativedelta import relativedelta  t a   datetime now   t b   datetime now    def diff t a  t b       t diff   relativedelta t b  t a     later end time comes first      return   h h  m m  s s  format h t diff hours  m t diff minutes  s t diff seconds    Regarding to the question you still need to use datetime strptime   as others said earlier

User · Answer

Structure that represent time difference in Python is called timedelta  If you have start time and end time as datetime types you can calculate the difference using - operator like   diff   end time - start time   you should do this before converting to particualr string format  eg  before start time strftime        In case you have already string representation you need to convert it back to time datetime by using strptime method

User · Answer

This site says to try   import datetime as dt start  09 35 23  end  10 23 00  start dt   dt datetime strptime start    H  M  S   end dt   dt datetime strptime end    H  M  S   diff    end dt - start dt   diff seconds 60    This forum uses time mktime

User · Answer

Both time and datetime have a date component   Normally if you are just dealing with the time part you d supply a default date  If you are just interested in the difference and know that both times are on the same day then construct a datetime for each with the day set to today and subtract the start from the stop time to get the interval  timedelta

User · Answer

Yes  definitely datetime is what you need here  Specifically  the strptime function  which parses a string into a time object   from datetime import datetime s1    10 33 26  s2    11 15 49    for example FMT     H  M  S  tdelta   datetime strptime s2  FMT  - datetime strptime s1  FMT    That gets you a timedelta object that contains the difference between the two times  You can do whatever you want with that  e g  converting it to seconds or adding it to another datetime   This will return a negative result if the end time is earlier than the start time  for example s1   12 00 00 and s2   05 00 00  If you want the code to assume the interval crosses midnight in this case  i e  it should assume the end time is never earlier than the start time   you can add the following lines to the above code   if tdelta days  lt  0      tdelta   timedelta days 0                  seconds tdelta seconds  microseconds tdelta microseconds     of course you need to include from datetime import timedelta somewhere   Thanks to J F  Sebastian for pointing out this use case

[python] How to calculate the time interval between two time strings

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