Delete rows from a table using a condition on datetimes

Hello!

I'm working with a table T having two separate datetime columns containing dates ('M\d\yyyy'): let' say the first column is called dates1 and the second is called dates2. I want to find a way to keep ONLY the rows where the column dates2 is one day ahead of column dates1.

Can something like this work? I'm dealing with a huge set of data and I don't want to make mistakes

indexes=(T.dates1+1)~=T.dates2; %Positions to delete
T(indexes,:)=[]; %delete all the columns in those positions

Hope that the question is clear.

Thank you!

Answer 1

[Steven Lord]

For the sake of humans reading your code I'd be more explicit and give that 1 some units.

rng default
t = datetime('today')

t = datetime
   27-Jan-2022

v1 = t + caldays(randi([-7 7], 1, 5))

v1 = 1×5 datetime array
   01-Feb-2022   02-Feb-2022   21-Jan-2022   02-Feb-2022   29-Jan-2022

v2 = v1 + caldays(randi([-1 1], 1, 5))

v2 = 1×5 datetime array
   31-Jan-2022   01-Feb-2022   21-Jan-2022   03-Feb-2022   30-Jan-2022

The fourth and fifth elements of v2 are 1 day after the corresponding elements of v1.

isOneDayLater = v2 == (v1 + caldays(1))

isOneDayLater = 1×5 logical array
   0   0   0   1   1

Why do I use caldays instead of just days?

daylightSavingsTimeStarts = datetime(2022, 3, 13, 'TimeZone', 'America/New_York')

daylightSavingsTimeStarts = datetime
   13-Mar-2022

daylightSavingsTimeStarts + days(1) % a 23 hour long day

ans = datetime
   14-Mar-2022 01:00:00

daylightSavingsTimeStarts + caldays(1)

ans = datetime
   14-Mar-2022

days(1) represents 24 hours. caldays(1) represents one calendar day, no matter how many hours that calendar day contains. That could be 23 hours, 24 hours, or 25 hours.

datetime(2022, 11, 6, 'TimeZone', 'America/New_York') + days(1) % DST ends

ans = datetime
   06-Nov-2022 23:00:00