Average time for datetime list

Question

Looking for fastest solution of time averaging problem.

I\'ve got a list of datetime objects. Need to find average value of time (excluding year, month, day). Here i

Answer 1

You would at least use sum() with a generator expression to create the total number of seconds:

from datetime import datetime, date, time

def avg_time(datetimes):
    total = sum(dt.hour * 3600 + dt.minute * 60 + dt.second for dt in datetimes)
    avg = total / len(datetimes)
    minutes, seconds = divmod(int(avg), 60)
    hours, minutes = divmod(minutes, 60)
    return datetime.combine(date(1900, 1, 1), time(hours, minutes, seconds))

Demo:

>>> from datetime import datetime, date, time, timedelta
>>> def avg_time(datetimes):
...     total = sum(dt.hour * 3600 + dt.minute * 60 + dt.second for dt in datetimes)
...     avg = total / len(datetimes)
...     minutes, seconds = divmod(int(avg), 60)
...     hours, minutes = divmod(minutes, 60)
...     return datetime.combine(date(1900, 1, 1), time(hours, minutes, seconds))
... 
>>> avg_time([datetime.now(), datetime.now() - timedelta(hours=12)])
datetime.datetime(1900, 1, 1, 7, 13)

Answer 2

Here's a short and sweet solution (perhaps not the fastest though). It takes the difference between each date in the date list and some arbitrary reference date (returning a datetime.timedelta), and then sums these differences and averages them. Then it adds back in the original reference date.

import datetime
def avg(dates):
  any_reference_date = datetime.datetime(1900, 1, 1)
  return any_reference_date + sum([date - any_reference_date for date in dates], datetime.timedelta()) / len(dates)

Answer 3

Here's a better way to approach this problem

Generate a sample of datetimes

In [28]: i = date_range('20130101',periods=20000000,freq='s')

In [29]: i
Out[29]: 
<class 'pandas.tseries.index.DatetimeIndex'>
[2013-01-01 00:00:00, ..., 2013-08-20 11:33:19]
Length: 20000000, Freq: S, Timezone: None

avg 20m times

In [30]: %timeit pd.to_timedelta(int((i.hour*3600+i.minute*60+i.second).mean()),unit='s')
1 loops, best of 3: 2.87 s per loop

The result as a timedelta (note that this requires numpy 1.7 and pandas 0.13 for the to_timedelta part, coming very soon)

In [31]: pd.to_timedelta(int((i.hour*3600+i.minute*60+i.second).mean()),unit='s')
Out[31]: 
0   11:59:12
dtype: timedelta64[ns]

In seconds (this will work for pandas 0.12, numpy >= 1.6).

In [32]: int((i.hour*3600+i.minute*60+i.second).mean())
Out[32]: 43152

Answer 4

I was looking for the same, but then i discovered this. A very simple way to get average of datetime object's list.

    import datetime
    #from datetime.datetime import timestamp,fromtimestamp,strftime ----> You can use this as well to remove unnecessary datetime.datetime prefix :)  
    def easyAverage(datetimeList): ----> Func Declaration
        sumOfTime=sum(map(datetime.datetime.timestamp,datetimeList))
        '''
         timestamp function changes the datetime object to a unix timestamp sort of a format.
         So I have used here a map to just change all the datetime object into a unix time stamp form , added them using sum and store them into sum variable.
        '''
        length=len(datetimeList) #----> Self Explanatory

        averageTimeInTimeStampFormat=datetime.datetime.fromtimestamp(sumOfTime/length)
        '''
        fromtimestamp function returns a datetime object from a unix timestamp.
        '''

        timeInHumanReadableForm=datetime.datetime.strftime(averageTimeInTimeStampFormat,"%H:%M:%S") #----> strftime to change the datetime object to string.
        return timeInHumanReadableForm

Or you can do all this in one simple line:

    avgTime=datetime.datetime.strftime(datetime.datetime.fromtimestamp(sum(map(datetime.datetime.timestamp,datetimeList))/len(datetimeList)),"%H:%M:%S")

Cheers,