Delete rows from a table using a condition on datetimes
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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!
采纳的回答
For the sake of humans reading your code I'd be more explicit and give that 1 some units.
rng default
t = datetime('today')
v1 = t + caldays(randi([-7 7], 1, 5))
v2 = v1 + caldays(randi([-1 1], 1, 5))
The fourth and fifth elements of v2 are 1 day after the corresponding elements of v1.
isOneDayLater = v2 == (v1 + caldays(1))
Why do I use caldays instead of just days?
daylightSavingsTimeStarts = datetime(2022, 3, 13, 'TimeZone', 'America/New_York')
daylightSavingsTimeStarts + days(1) % a 23 hour long day
daylightSavingsTimeStarts + caldays(1)
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
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