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 is what I got so far:
import datetime as dtm
def avg_time(times):
avg = 0
for elem in times:
avg += elem.second + 60*elem.minute + 3600*elem.hour
avg /= len(times)
rez = str(avg/3600) + ' ' + str((avg%3600)/60) + ' ' + str(avg%60)
return dtm.datetime.strptime(rez, "%H %M %S")
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
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)
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,
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)
