Logical indexing for matrix
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Jakob Hannibal
il 21 Nov 2014
Commentato: Star Strider
il 27 Nov 2014
Hi Everybody! I have difficulties with some logical indexing.
I have vector with time values. For example the date 1st january 2008 at 9:00:00 am would be
tvec=2008 1 1 9 0 0 and so on...
Tvec covers a whole year so it has 527040 rows and 6 columns.
How do I locate a specic period. For example 1st january 2008 from 9 to 10 am?
I tried this:
clear;clc;
period = [2008 1 1 9];
idx(:,1)=tvec(:,1)==period(1);
idx(:,2)=tvec(:,2)==period(2);
idx(:,3)=tvec(:,3)==period(3);
idx(:,4)=tvec(:,4)==period(4);
L=logical(idx);
tvec_a=tvec(L);
Thanks for any help you might have...
Risposta accettata
Star Strider
il 21 Nov 2014
I’m not sure logical indexing is necessary. The find function will probably be preferable:
dn = datenum([2008 01 01 0 0 0]); % Create Data
dnv = dn:(1/24):(dn+2); % Create Data
dv = datevec(dnv);
time_start = datenum([2008 01 01 09 00 00]);
time_end = datenum([2008 01 01 10 00 00]);
rngidx = find( (dnv>=time_start) & (dnv<=time_end));
out = dv(rngidx,:)
9 Commenti
Star Strider
il 27 Nov 2014
I’m lost. I still don’t know if you’re supposed to aggregate particular months across all years, particular months in each given year, or what. Using find first on the years and then on the months (or days or whatever you decide) will work. My problem is that I don’t know in detail what you want to do for your project.
Più risposte (1)
per isakson
il 21 Nov 2014
Modificato: per isakson
il 21 Nov 2014
Replace
L=logical(idx);
by
L=logical( all(idx,2));
idx is already logical, no need apply the function, logical.
 
The datevec-format is not appropriate for this task when it comes to longer periods. Convert to serial date number.
sdn = datenum( tvec );
sdn1 = datenum( [2008,1,1, 9,0,0] );
sdn2 = datenum( [2008,1,1,10,0,0] );
Here is a problem with numerical precision. I prefer to carefully convert to seconds (whole numbers) to make comparisons simpler.
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