Extract rows from matrix based on date

I have a 88416 x 11 matrix. Hourly date and time datenum values are in column 1. The hourly date range is from 2006-12-1 to 2016-12-31. If the date is 2007-10-28 I want to remove it from the matrix and create a new matrix that contains it and all the values in that row. I want to repeat this for several dates. How do I do it?

Answer 1

[Walter Roberson]

If you have R2016b or later, use a timetable() object, and extract rows using a timerange; https://www.mathworks.com/help/matlab/matlab_prog/subscript-into-times-of-timetable.html#zmw57dd0e28127

[David du Preez]

I have a older version of MATLAB. Is it possible to do what I want to do.

[Walter Roberson]

dv = datevec(YourArray(:,1));

match = dv(:,1) == 2007 & dv(:,2) == 10 && dv(:,3) == 28;
extracted_data = YourArray(match,:);

or:

match = floor(YourArray(:,1)) == datenum([2007, 10, 28]);
extracted_data = YourArray(match,:);

[David du Preez]

Awesome! thank you.

How would it work for a range of dates from 2007-10-28 to 2007-11-10?

[Walter Roberson]

fd = floor(YourArray(:,1));
match = fd >= datenum([2007,10,28]) & fd <= datenum([2007,11,10]);
extracted_data = YourArray(match,:);

Alternately,

match = YourArray(:,1) >= datenum([2007,10,28]) & YourArray(:,1) < (datenum([2007,11,10]) + 1);
extracted_data = YourArray(match, :);

Note that the first operation is >= but the second operation is strictly < comparing to one day after your last day. This is because datenum are in whole days and fractions of a day, so everything up to .999999999 (etc) of the ending day belongs to the ending day, as soon as you get to the whole number next day the date stops belonging to the range.

Note: this end date calculation is not guaranteed to be valid for the days that have leap seconds. datenum format is quite weak in the handling of leap seconds.

Answer 2

[Peter Perkins]

David, even prior to R2016b, you might look at using datetimes rather than datenums:

>> d = datetime(2017,10,26,(0:60:420)',0,0)
d = 
  8×1 datetime array
   26-Oct-2017 00:00:00
   28-Oct-2017 12:00:00
   31-Oct-2017 00:00:00
   02-Nov-2017 12:00:00
   05-Nov-2017 00:00:00
   07-Nov-2017 12:00:00
   10-Nov-2017 00:00:00
   12-Nov-2017 12:00:00
>> ( '28-Oct-2017'<= d & d<'11-Nov-2017' )'
ans =
  1×8 logical array
   0   1   1   1   1   1   1   0

Put d and the rest of your data in a table, and although you don't get everything that timetables provide, you might find it easier to subset your data.