subsetting dates in a matrix

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Cagdas Ozgenc
Cagdas Ozgenc on 1 Nov 2013
Commented: Cedric on 1 Nov 2013
Hello,
I have a time series in a matrix and first column is serial date numbers (up to millisecond precision), and other columns my samples. I would like to get those rows that belong to for example '12-Oct-2012'. I couldn't figure out how to write the logical operator to get a slice of this matrix. I would like to hopefully avoid some sort of slow string comparison.
Thanks in advance
  2 Comments
Azzi Abdelmalek
Azzi Abdelmalek on 1 Nov 2013
Show the two first rows of your data
Cagdas Ozgenc
Cagdas Ozgenc on 1 Nov 2013
Edited: Cagdas Ozgenc on 1 Nov 2013
735236.4590277778 1.4264e+03 1.4272e+03 1.4217e+03
735236.4597222222 1.4237e+03 1.4252e+03 1.4234e+03
735236.4604166667 1.4247e+03 1.4267e+03 1.4244e+03

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Accepted Answer

Cedric
Cedric on 1 Nov 2013
Edited: Cedric on 1 Nov 2013
You already have serialized/numeric time stamps in you array, so just convert date boundaries to numeric, and compare numbers: assuming your array is named data..
lb_incl = datenum( '12-Oct-2012', 'dd-mmm-yyyy' ) ;
ub_excl = datenum( '13-Oct-2012', 'dd-mmm-yyyy' ) ;
id = data(:,1) >= lb_incl & data(:,1) < ub_excl ;
dailyMean = mean( data(id,2) ) ;
or
dailySlice = data(id,2:end) ;
  2 Comments
Cagdas Ozgenc
Cagdas Ozgenc on 1 Nov 2013
Thanks. In this scenario this is a good option.
Cedric
Cedric on 1 Nov 2013
You're welcome.

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More Answers (1)

Azzi Abdelmalek
Azzi Abdelmalek on 1 Nov 2013
If A is your cell array
c1=cellstr(datestr(A(:,1),'dd-mm-yyyy'))
idx=ismember(c1,'12-10-2012')
out=A(idx,:)
  1 Comment
Cagdas Ozgenc
Cagdas Ozgenc on 1 Nov 2013
Takes forever on my dataset of million lines. It converts all serial numbers back to string and then compares those strings. Then gets the indices. Then gets the data.
Thanks anyways

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