How to shift date-time variable to a common start-date/time

I have some data formatted as follows:

% Simple example as the "actual_dates" is a large char array.
common_date = datetime(14,1,1,0,0,0) % Define a common start date (01-Jan-0014 00:00:00)
actual_dates = '29-Jan-2014 13:57:30' % This is "char"

Then I run this:

t_diff = actual_dates - common_date % The shift value

and get this weird result:

t_diff = 17532325:57:35

The end goal is to have this:

shifted_date = actual_dates - t_diff

For this particular example, want to have shifted_date = '01-Jan-0014 00:00:00'

How can I fix this? I'm not sure how to convert the 'char' array into a datetime format.

Answer 1

[Star Strider]

I don’t understand what you want to do, but converting the string to a datetime object is srtaightforward. Doing calculations on them is a bit more involved, since it requires the appropriate functions.

Try this to start:

common_date = datetime(14,1,1,0,0,0);
actual_dates = '29-Jan-2014 13:57:30';
act_date = datetime(actual_dates);
t_diff = between(common_date,act_date)

produces:

t_diff = 
   2000y 28d 13h 57m 30s

[David O'Brien]

All I needed to do was to shift multiple sets of recorded dates/times to all start at a certain arbitrary time/date.

Great. Thanks for your help! The between function and using datetime were just what I needed.

[Star Strider]

As always, my pleasure!

Answer 2

[Katalin]

Try this: shifted_date = datetime(actual_dates) - t_diff

[David O'Brien]

Thanks! That helped as well.

Answer 3

[Peter Perkins]

Just for the record, this

>> common_date = datetime(14,1,1,0,0,0)
common_date = 
   01-Jan-0014 00:00:00
>> actual_dates = '29-Jan-2014 13:57:30'
actual_dates =
29-Jan-2014 13:57:30
>> t_diff = actual_dates - common_date
t_diff = 
   17532325:57:30

does automatically convert the string (no need to explicitly call datetime on it, although of course it doesn't hurt). It seems a little funny that you are creating a datetime in the year 14 (not 2014), but I'll assume you meant to do that. The result simply says that there are 17.5 million hours between those two dates. This

>> t_diff.Format = 'y'
t_diff = 
   2000.1 yrs

is probably a more useful format, or, as StarStrider suggested, use the between function to get the difference in calendar units.