Given a string like this:
2020-04-09T13:15:52.838Z - 2020-04-09T13:16:09.704Z
What is the correct way to calculate the amount of elapsed seconds? So far I tried this:
s = '2021-03-09T13:15:52.838Z - 2021-03-09T13:16:09.704Z'.split(' - ')
datetime.strptime(s[0], "%Y-%m-%d").date() - datetime.strptime(s[1],"%Y-%m-%d").date()
However, I am not sure how to extract the seconds and calculate the elapsed time between the two time stamps. Any idea of how to get the elapsed time in seconds?
Basically, you want to convert each string into a datetime (here using datetime.strptime() and using a format string that matches the format of your time strings) and then subtract one from the other to get a datetime.timedelta object. Then you can call the total_seconds() method on the timedelta object. Here's one way you might do it:
import datetime
s = '2021-03-09T13:15:52.838Z - 2021-03-09T13:16:09.704Z'.split(' - ')
first_time = datetime.datetime.strptime(s[0], '%Y-%m-%dT%H:%M:%S.%fZ')
second_time = datetime.datetime.strptime(s[1], '%Y-%m-%dT%H:%M:%S.%fZ')
time_delta = second_time - first_time
print(time_delta.total_seconds())
I have a timestep (e.g. 2717715 microseconds) which I want to convert to hh:mm:ss format to extract a frame from a youtube video. I have used the following code to do so:
def convert(timestamp):
# pdb.set_trace()
timestamp = float(timestamp)
# seconds = (micros/1000)%60
# minutes = (micros/(1000*60))%60
# hours = (micros/(1000*60*60))%24
# pdb.set_trace
milliseconds = (timestamp / 1000) % 1000
seconds = (( (timestamp / 1000) - milliseconds)/1000)%60
minutes = (((( (timestamp / 1000) - milliseconds)/1000) - seconds)/60) %60
hours = ((((((timestamp / 1000) - milliseconds)/1000) - seconds)/60) - minutes)/60
return hours, minutes, seconds
But it is not working. Can you please tell me what is wrong with my code?
Considering your timestamp is of type float you can use the built in divmod function as follows:
timestamp = 3 * 3600 + 7 * 60 + 20 # just creating (3h 7m 20s) float format timestamp
print(timestamp) # gives 11240
suppose you have that timestamp, you can extract your hours, minutes and seconds from it as follows:
timestamp = float(timestamp)/1000000 # The timestamp is in microseconds
hours,remainder = divmod(timestamp, 3600)
minutes,seconds = divmod(remainder, 60)
print(hours) # gives 3
print(minutes) # gives 7
print(seconds) # gives 20
Hope that helped. Happy Coding.
I think you are looking for the datetime module.
You can get a human-readable time from a timestamp in the following manner:
from datetime import datetime
def convert(timestamp):
return datetime.fromtimestamp(timestamp).time()
I want to calculate difference between two time in hours using django in sql db the time are stored in timefield.
I tried this:
def DesigInfo(request): # attendance summary
emplist = models.staff.objects.values('empId', 'name')
fDate = request.POST.get('fromDate')
tDate = request.POST.get('toDate')
if request.GET.get('empId_id'):
sel = attendance.objects.filter(empId_id=request.GET.get('empId_id'),)
for i in sel:
# print i.
# print i.outTime
# print i.inTime.hour,i.inTime.minute,i.inTime.second - i.outTime.hour,i.outTime.minute,i.outTime.second
ss = i.inTime.hour
ss1 = 12 - ss
mm = i.outTime.hour
mm1 = (12 + mm) - 12
print ss1 + mm1
Since i.inTime and i.outTime are time objects you cannot simply subtract them. A good approach is to convert them to datetime adding the date part (use today() but it is irrelevant to the difference), then subtract obtaining a timedelta object.
delta = datetime.combine(date.today(), i.outTime) - datetime.combine(date.today(), i.inTime)
(Look here: subtract two times in python)
Then if you want to express delta in hours:
delta_hours = delta.days * 24 + delta.seconds / 3600.0
A timedelta object has 3 properties representing 3 different resolutions for time differences (days, seconds and microseconds). In the last expression I avoided to add the microseconds but I suppose it is not relevant in your case. If it is also add delta.microseconds / 3600000000.0
Note that simply dividing seconds by 3600 would have returned only the integer part of hours avoiding fractions. It depends on your business rules how to round it up (round, floor, ceil or leave the fractional part as I did)
Using datetime objects: https://docs.python.org/2/library/datetime.html
A good stack overflow post on the topic How to get current time in Python
from datetime import datetime
now = datetime.now()
# wait some time
then = ... some time
# diff is a datetime.timedelta instance
diff = then - now
diff_hours = diff.seconds / 3600
You might want to play with this codes:
from datetime import datetime
#set the date and time format
date_format = "%m-%d-%Y %H:%M:%S"
#convert string to actual date and time
time1 = datetime.strptime('8-01-2008 00:00:00', date_format)
time2 = datetime.strptime('8-02-2008 01:30:00', date_format)
#find the difference between two dates
diff = time2 - time1
''' days and overall hours between two dates '''
print ('Days & Overall hours from the above two dates')
#print days
days = diff.days
print (str(days) + ' day(s)')
#print overall hours
days_to_hours = days * 24
diff_btw_two_times = (diff.seconds) / 3600
overall_hours = days_to_hours + diff_btw_two_times
print (str(overall_hours) + ' hours');
''' now print only the time difference '''
''' between two times (date is ignored) '''
print ('\nTime difference between two times (date is not considered)')
#like days there is no hours in python
#but it has seconds, finding hours from seconds is easy
#just divide it by 3600
hours = (diff.seconds) / 3600
print (str(hours) + ' Hours')
#same for minutes just divide the seconds by 60
minutes = (diff.seconds) / 60
print (str(minutes) + ' Minutes')
#to print seconds, you know already ;)
print (str(diff.seconds) + ' secs')
The easiest way through I achieve is the comment of Zac given above. I was using relativedelta like this
from dateutil import relativedelta
difference = relativedelta.relativedelta( date1, date2)
no_of_hours = difference.hours
but it did not give me correct result when the days changes. So, I used the approach expressed above:
no_of_hours = (difference.days * 24) + (difference.seconds / 3600)
Please note that you will be getting negative value if date2 is greater than date1. So, you need to swipe the position of date variables in relativedelta.
I have a timedelta object which is 3457 hours
timedelta(hours=3457)
I want to represent it in "HH:MM" format, which is "3457:00"
I do:
from datetime import datetime
hours = timedelta(hours=3457)
hours_string = time.strftime("%H:%M", time.gmtime(hours.seconds))
print hours_string
"01:00"
How can I get "3457:00"?
Please note that 3457:00 is a nonsensical format. The "hour-colon-minutes" format is used in dates and times, and the hour then can't reasonably be any higher than 23. A more reasonable format is: 3457h 0m.
You can get it like this:
from datetime import timedelta
delta = timedelta(hours=3457)
minutes, seconds = divmod(delta.seconds, 60)
hours, minutes = divmod(minutes, 60)
hours += delta.days * 24
print '%sh %sm' % (hours, minutes)
Of course, an easier way is this:
from datetime import timedelta
delta = timedelta(hours=3457)
print delta
But that will give you "144 days, 1:00:00", which is a sane format, but not what you want.
As the timedelta documentation notes, only days, seconds and microseconds are stored internally -- which means you'll have to manually convert them to the units you want (hours and minutes). Here's one way to do that:
from datetime import timedelta
d = timedelta(hours=3457, minutes=42)
wholehours, seconds = divmod(d.seconds, 60*60)
wholeminutes = seconds//60
deltahours = d.days*24 + wholehours
print('{:d}:{:02d}'.format(deltahours, wholeminutes))
# 3457:42
Here's a simpler alternative that produces the same result:
def deltatime_hours_mins(dt, sep=':'):
secs = int(dt.total_seconds())
return '{:d}{}{:02d}'.format(secs//3600, sep, secs//60 % 60)
print(deltatime_hours_mins(d))
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?
Yes, definitely datetime is what you need here. Specifically, the datetime.strptime() method, which parses a string into a datetime 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 < 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.
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.
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 <= end: # e.g., 10:33:26-11:15:49
return end - start
else: # end < start e.g., 23:55:00-00:25:00
end += timedelta(1) # +day
assert end > 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)))
# -> 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.
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()
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.
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.
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
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
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).
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.
you can use pendulum:
import pendulum
t1 = pendulum.parse("10:33:26")
t2 = pendulum.parse("10:43:36")
period = t2 - t1
print(period.seconds)
would output:
610
import datetime
day = int(input("day[1,2,3,..31]: "))
month = int(input("Month[1,2,3,...12]: "))
year = int(input("year[0~2020]: "))
start_date = datetime.date(year, month, day)
day = int(input("day[1,2,3,..31]: "))
month = int(input("Month[1,2,3,...12]: "))
year = int(input("year[0~2020]: "))
end_date = datetime.date(year, month, day)
time_difference = end_date - start_date
age = time_difference.days
print("Total days: " + str(age))
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)
Usually, you have more than one case to deal with and perhaps have it in a pd.DataFrame(data) format. Then:
import pandas as pd
df['duration'] = pd.to_datetime(df['stop time']) - pd.to_datetime(df['start time'])
gives you the time difference without any manual conversion.
Taken from Convert DataFrame column type from string to datetime.
If you are lazy and do not mind the overhead of pandas, then you could do this even for just one entry.
Here is the code if the string contains days also [-1 day 32:43:02]:
print(
(int(time.replace('-', '').split(' ')[0]) * 24) * 60
+ (int(time.split(' ')[-1].split(':')[0]) * 60)
+ int(time.split(' ')[-1].split(':')[1])
)