How do I calculate the difference in time in minutes for the following timestamp in Python?
2010-01-01 17:31:22
2010-01-03 17:31:22
there is also a sneak way with pandas:
pd.to_timedelta(x) - pd.to_timedelta(y)
from datetime import datetime
fmt = '%Y-%m-%d %H:%M:%S'
d1 = datetime.strptime('2010-01-01 17:31:22', fmt)
d2 = datetime.strptime('2010-01-03 17:31:22', fmt)
print (d2-d1).days * 24 * 60
In Other ways to get difference between date;
import dateutil.parser
import datetime
timeDifference = current_date - dateutil.parser.parse(last_sent_date)
time_difference_in_minutes = (int(timeDifference.days) * 24 * 60) + int((timeDifference.seconds) / 60)
Thanks
minutes_diff = (datetime_end - datetime_start).total_seconds() / 60.0
RSabet's answer doesn't work in cases where the dates don't have the same exact time.
Original problem:
from datetime import datetime
fmt = '%Y-%m-%d %H:%M:%S'
d1 = datetime.strptime('2010-01-01 17:31:22', fmt)
d2 = datetime.strptime('2010-01-03 17:31:22', fmt)
daysDiff = (d2-d1).days
print daysDiff
> 2
# Convert days to minutes
minutesDiff = daysDiff * 24 * 60
print minutesDiff
> 2880
d2-d1 gives you a datetime.timedelta and when you use days it will only show you the days in the timedelta. In this case it works fine, but if you would have the following.
from datetime import datetime
fmt = '%Y-%m-%d %H:%M:%S'
d1 = datetime.strptime('2010-01-01 16:31:22', fmt)
d2 = datetime.strptime('2010-01-03 20:15:14', fmt)
daysDiff = (d2-d1).days
print daysDiff
> 2
# Convert days to minutes
minutesDiff = daysDiff * 24 * 60
print minutesDiff
> 2880 # that is wrong
It would have still given you the same answer since it still returns 2 for days; it ignores the hour, min and second from the timedelta.
A better approach would be to convert the dates to a common format and then do the calculation. The easiest way to do this is to convert them to Unix timestamps. Here is the code to do that.
from datetime import datetime
import time
fmt = '%Y-%m-%d %H:%M:%S'
d1 = datetime.strptime('2010-01-01 17:31:22', fmt)
d2 = datetime.strptime('2010-01-03 20:15:14', fmt)
# Convert to Unix timestamp
d1_ts = time.mktime(d1.timetuple())
d2_ts = time.mktime(d2.timetuple())
# They are now in seconds, subtract and then divide by 60 to get minutes.
print int(d2_ts-d1_ts) / 60
> 3043 # Much better
The result depends on the timezone that corresponds to the input time strings.
The simplest case if both dates use the same utc offset:
#!/usr/bin/env python3
from datetime import datetime, timedelta
time_format = "%Y-%d-%m %H:%M:%S"
dt1 = datetime.strptime("2010-01-01 17:31:22", time_format)
dt2 = datetime.strptime("2010-01-03 17:31:22", time_format)
print((dt2 - dt1) // timedelta(minutes=1)) # minutes
If your Python version doesn't support td // timedelta
; replace it with int(td.total_seconds() // 60)
.
If the input time is in the local timezone that might have different utc offset at different times e.g., it has daylight saving time then you should make dt1
, dt2
into aware datetime objects before finding the difference, to take into account the possible changes in the utc offset.
The portable way to make an aware local datetime objects is to use pytz
timezones:
#!/usr/bin/env python
from datetime import timedelta
import tzlocal # $ pip install tzlocal
local_tz = tzlocal.get_localzone() # get pytz timezone
aware_dt1, aware_dt2 = map(local_tz.localize, [dt1, dt2])
td = aware_dt2 - aware_dt1 # elapsed time
If either dt1
or dt2
correspond to an ambiguous time then the default is_dst=False
is used to disambiguate. You could set is_dst=None
to raise an exception for ambiguous or non-existent local times instead.
If you can't install 3rd party modules then time.mktime() could be used from @Ken Cochrane's answer that can find the correct utc offset on some platforms for some dates in some timezones -- if you don't need a consistent (but perhaps wrong) result then it is much better than doing dt2 - dt1
with naive datetime objects that always fails if the corresponding utc offsets are different.