converting and comparing 24hr cell to a minimum value

Hoping to convert a string called "adjTIB" from a 24 hour cell ie. 14:11 to compare whether it is > 21:00, and < 09:00

I assume I need to run a cellfun to compare inequality to a constant, but not sure if there is an easier way.

Answer 1

[Adam Danz]

"Hoping to convert a string called "adjTIB" from a 24 hour cell ie. 14:11 "

This is a bit ambiguous. Does the cell array contain strings? Datetime values? Durations? If what format? Provide an example or detailed description and I'd be glad to help out with that step.

" compare whether [the values are] > 21:00, and < 09:00 "

A time or duration cannot be greater than 21:00 and less than 9:00 at the same time. I'm guessing that you want to identify times or durations that are less than 9:00 or greater than 21:00. First convert the values in your cell array to either durations or datetime. This isn't an arbitrary choice; chose the format that describe your data.

Here's how to do that with Datetime values

d = datetime(1999,03,16):minutes(15):datetime(1999,03,17); 
idx = hour(d) > 21 | hour(d) < 9;

... or Durations

d = hours(rand(1,20)*24);
idx = d > hours(21) | d < hours(9);

[Noah Milman]

You're correct, I did mean greater than 9 pm and earlier than 9 am, looking for the indices that include values within that range. I apologize

Originally, the data is in the form of a datetime variable to which I converted using datestr

datestr(InBedTime(i,5),15)
% converts to an output of HH:MM in 24Hr format

But looking at your advice, its much simpler to handle and manipulate the original datetime variable. Using your code, I can identify the indices that fall within the 9 pm - 9 am range, and analyze that data.

thanks for your help.

[Adam Danz]

" its much simpler to handle and manipulate the original datetime variable"

Yes! That will almost always be the case. I take it the problem is solved, then. Let me know if there are any leftover problems.