Determine if a day is a bank holiday (from a matrix)

Hello there,

I have an issue, trying to check if a datetime vector is a bank holiday. The datetimes of the bankholidays are included in a seperate vector, called BankHolidays.

   BankHolidays = 
   8×1 datetime array
   02-01-2017
   14-04-2017
   17-04-2017
   01-05-2017
   29-05-2017
   28-08-2017
   25-12-2017
   26-12-2017

How can I check if a specific datetime vector (e.g. date1) is a Bankholiday by comparing its datetime with the datetime of BankHolidays?

I have tried this:

 any(date1 == BankHolidays)  == false

But the results are not good. I also tried to calculate the respective datenum of the two vectors but I am getting different results for the same days, despite the fact that the date format is the same (dd/mm/yyyy).

I would appreciate your insight into this.

PS. I would like to avoid isbusday.

Thank you in advance,

Andreas

Answer 1

[Fangjun Jiang]

t1 = datetime(2014,1:5,1)
t2 = datetime(2014,1,1)
ismember(t2,t1)

[Andreas Georgakakos]

Thank you so much. Because my datetime vector includes hours (e.g. 1 Jan 2017 15:00) I am getting bad results for some of the values. Do you happen to know how I can remove the hours and seconds from my datetime vector so that it will just be 1 Jan 2017)?

Thank you.

[Steven Lord]

Depending on the exact datetime values stored in date1 (whether they're values representing midnight or if they include time information) you may also need to dateshift to remove hours, minutes, and seconds.

>> NOW = datetime('now');
>> THEN = datetime('today');
>> ismember(NOW, THEN)
ans =
  logical
   0
>> soon = ismember(dateshift(NOW, 'start', 'day'), THEN)
soon =
  logical
   1

The variables NOW and THEN aren't the same, since I'm not running this code at the stroke of midnight. But they are the same day, so when I shift NOW back to the start of the day it is equal to THEN. If you wanted to be extra careful and avoid the possibility of NOW being created immediately before midnight and THEN immediately after, you could dateshift to the start of the day that is 'nearest'. See the dateshift documentation for more information.

I could have used a different variable name than now instead of writing it in upper case if I wanted to avoid shadowing the now function, but the Spaceballs reference wouldn't have worked if I had. :)

[Fangjun Jiang]

You can also use t=datetime('now','format','dd-MM-uuuu')

[Peter Perkins]

Unless one is comparing text (which is not the best way to do this), that's not correct, Fangjun. Changing the display format has no effect at all on the value:

>>  t = datetime('now','format','dd-MM-uuuu')
t = 
  datetime
   13-02-2018
>> timeofday(t)
ans = 
  duration
   15:06:00

[Fangjun Jiang]

Got it. Thanks Peter!

[Andreas Georgakakos]

Hi Steven,

My datetimes are hourly for a whole year (8760 values), for example:

01/01/2017 00:00

01/01/2017 01:00 ....

31/12/2017 23:00

In order to compare dates either by checking equality or by converting to datenum (?), is it better to remove hours and minutes or converting all the hours to 00:00 ?

Thanks.

Andreas

[Fangjun Jiang]

t=datetime(2017,1,1):hours(1):datetime(2017,12,31,23,00,00)
t2=dateshift(t,'start','day')
ismember(t2,BankHolidays)