subtract two times in python

Question

I have two datetime time values  exit and enter and I want to do something like   duration   exit - enter   However  I get this error      TypeError  unsupported operand type s  for -   datetime time  and    datetime time   How do I do this correctly  One possible solution is converting the time variables to datetime variables and then subtruct  but I m sure you guys must have a better and cleaner way

User · Answer

Try this   from datetime import datetime  date  datetime combine date today    exit  - datetime combine date today    enter    combine builds a datetime  that can be subtracted

User · Answer

Use   from datetime import datetime  date  duration   datetime combine date min  end  - datetime combine date min  beginning    Using date min is a bit more concise and works even at midnight   This might not be the case with date today   that might return unexpected results if the first call happens at 23 59 59 and the next one at 00 00 00

User · Answer

You have two datetime time objects so for that you just create two timedelta using datetime timedetla and then substract as you do right now using  -  operand  Following is the example way to substract two times without using datetime   enter   datetime time hour 1     Example enter time exit   datetime time hour 2     Example start time enter delta   datetime timedelta hours enter hour  minutes enter minute  seconds enter second  exit delta   datetime timedelta hours exit hour  minutes exit minute  seconds exit second  difference delta   exit delta - enter delta   difference delta is your difference which you can use for your reasons

User · Answer

datetime time does not support this  because it s nigh meaningless to subtract times in this manner  Use a full datetime datetime if you want to do this

User · Answer

datetime time can not do it - But you could use datetime datetime now    start   datetime datetime now   sleep 10  end   datetime datetime now   duration   end - start

User · Answer

import datetime  def diff times in seconds t1  t2         caveat emptor - assumes t1  amp  t2 are python times  on the same day and t2 is after t1     h1  m1  s1   t1 hour  t1 minute  t1 second     h2  m2  s2   t2 hour  t2 minute  t2 second     t1 secs   s1   60    m1   60 h1      t2 secs   s2   60    m2   60 h2      return  t2 secs - t1 secs     using it diff times in seconds  datetime datetime strptime   13 23 34     H  M  S   time   datetime datetime strptime   14 02 39     H  M  S   time

User · Answer

I had similar situation as you and I ended up with using external library called arrow  Here is what it looks like   gt  gt  gt  import arrow  gt  gt  gt  enter   arrow get  12 30 45    HH mm ss    gt  gt  gt  exit   arrow now    gt  gt  gt  duration   exit - enter  gt  gt  gt  duration datetime timedelta 736225  14377  757451

User · Answer

The python timedelta library should do what you need  A timedelta is returned when you subtract two datetime instances   import datetime dt started   datetime datetime utcnow      do some stuff  dt ended   datetime datetime utcnow   print  dt ended - dt started  total seconds

User · Answer

timedelta accepts minus -  time values  so it could be simple as below import datetime  enter   datetime time hour 1  minute 30  exit   datetime time hour 2  minute 0  duration   datetime timedelta hours exit hour-enter hour  minutes exit minute-enter minute     duration   datetime timedelta hours 1  minutes -30   result  gt  gt  gt  duration datetime timedelta seconds 1800

User · Answer

instead of using time try timedelta   from datetime import timedelta  t1   timedelta hours 7  minutes 36  t2   timedelta hours 11  minutes 32  t3   timedelta hours 13  minutes 7  t4   timedelta hours 21  minutes 0   arrival   t2 - t1 lunch    t3 - t2 - timedelta hours 1   departure   t4 - t3  print arrival  lunch  departure

[python] subtract two times in python

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