Round date and hour to the previous 15 minute interval

Hello

I would like to round the datetime variable to the previous 15 minutes interval. The tricky part is that it has to be based on the current time. For instance:

t1 = datetime('now','Format','yyyy-MM-dd HH:mm')

the result of the previous code is

2018-03-08 10:34

So, I would like to obtain ...

2018-03-08 10:30

I have tried to used the dateshift function however, I haven't had a good solution

Answer 1

[Rik]

Dateshift would be a good solution, but it can only shift to quarter years, not quarter hours. The code below is a way around this.

t1 = datetime('now','Format','yyyy-MM-dd HH:mm')
t1.Minute = 15 * floor(t1.Minute/15)

%original code:
% t1 = datetime('now','Format','yyyy-MM-dd HH:mm');
% %syntax: t = datetime(Y,M,D,H,MI,S);
% t1=datetime(year(t1),month(t1),day(t1),hour(t1),15*floor(minute(t1)/15),0);

[Jose Valles]

It works!!!

Thanks!

[Jos (10584)]

No need to call date time twice:

t1 = datetime('now','Format','yyyy-MM-dd HH:mm')
t1.Minute = 15 * floor(t1.Minute/15)

[Rik]

Good point, I'll edit my answer.

[Jim McIntyre]

This answer works for the example given, but is incomplete for real times that might also have non-zero values for seconds. The answer rounds the minutes, but leaves the seconds unrounded.

To fully round to minutes, you need to add:

t1.Second = 0

Answer 2

[Steven Lord]

Let's start off with the current date and time.

N = datetime('now')

Figure out the start of the hour containing N.

H = dateshift(N, 'start', 'hour')

Generate a datetime vector spaced a quarter of an hour apart, starting with H.

Q = hours(0.25)
P = H + Q*(0:4)

If you're using release R2016b or later you can use discretize.

previousQuarterHour = P(discretize(N, P))

If you're using release R2015b or later you can interpolate using interp1.

previousQuarterHour = interp1(P, P, N, 'previous')

If you're using release R2014b or R2015a you'll need to use find to find the last element of P that is less than or equal to N.

previousQuarterHour = P(find(P <= N, 1, 'last'))

[Jose Valles]

Thanks!! it is also a good result

Answer 3

[Jose Valles]

Thank for your reply!!

One quick question always related to this post: if I want to find the previous 15 minutes, what would be the best way to do it?

For example, for a datetime variable like:

2018-03-08 08:06
2018-03-08 10:34

we want the last 15-minute as shown as

2018-03-08 07:15
2018-03-08 10:15

What would be the efficient way

Thank you so much!

[Rik]

Please don't use the answer field for comments.

What do you mean? I don't see the pattern. If you mean rounding to the next 15 minutes instead of the last, just replace floor with ceil.

[Jose Valles]

No. What I want is to round to previous 15 minutes (Not the previous 15 minutes interval). For instance, the datetime is '2018-03-08 08:06' so the result that I am looking for is '2018-03-08 07:15'.

Another example: the datetime is '2018-03-08 10:34' so the result that i am looking is '2018-03-08 10:15'

I just realized that my example from the previous post was innacurated. My apologies for the innacurate example

[Rik]

t1 = datetime('now','Format','yyyy-MM-dd HH:mm')
t1.Minute = 60 * floor((t1.Minute-15)/60)+15

This will round to the previous ##:15