以特定格式在python中计算经过时间

Question

I am trying to write a program to determine the time and date corresponding to a elapsed number of seconds since 00: 00: 00 on 1 January 2016.

But i wanted my result in specific format. ie should be the corresponding hour (in military time), minute, second, day of the month, month name, year, and day of the week name (Sunday – Saturday). for example,output should look like the following

23:59:32 2 January 2018 Tuesday 

i tried to code this

import time
start = time.time()
#do something
end = time.time()
temp = end-start
print(temp)
hours = temp//3600
temp = temp - 3600*hours
minutes = temp//60
seconds = temp - 60*minutes
print('%d:%d:%d' %(hours,minutes,seconds))

But i could only get the hours, minutes and seconds part, Is there any way i can get the month,year,and week name too ?

Also i'm trying to handle leap years too.since a leap year has 366 days with 29 days in February. Leap years are years that are evenly divisible by 4, with the exception of those evenly divisible by 100 but not 400.

Answer 1

To format a datetime you can use strftime().

import datetime
my_time = datetime.datetime(year=2018, month=1, day=2, hour=23, minute=59,second=32)
print (my_time.strftime("%X %-d %B %Y %A"))

If you want to change the format use this table for reference

Answer 2

You want to use the Datetime Module . Handling dates and times is trickier than it appears! That's why it's good that Python has datetime handling built-in. Datetime and timedelta objects account for leap years and leap seconds, and they handle operations like addition and subtraction 'intuitively'.

import datetime
def convert_seconds_since_jan_1(seconds):
    JAN_1_2001 = datetime.datetime(year = 2001, month = 1, day = 1)
    added_time = datetime.timedelta(seconds = seconds)    
    return (JAN_1_2001+added_time)

Getting the weekday is simply a matter of string formatting. Remember that weekdays are modulo 7!