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extracting rows based on string in first column (Should be fairly simple ... not for me apparently)

Asked by mashtine on 23 Apr 2014
Latest activity Edited by per isakson
on 23 Apr 2014
Hey everyone,
I have a large cell array with 8 columns (8758 x 8 cell). The first containing time in the string format of '2010-01-01 00:00'. I am trying to extract all the rows where the timestamp is '01' at the 9th and 10th characters.
I know this is relatively basic but cannot get the indexing right. I had this so far:
rows = find(a{:,1}(9,10) == '01')
extrdata = a{rows,1};
I have also tried the following but get the following error:
>> a( a{:,1}(9:10) == '01' , : )
Bad cell reference operation.
If I were to index just for one row then it works but my indexing is wrong to work for all rows:
>> a{1,1}(9:10)
ans =
01
Thanks for any help!

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1 Answer

Answer by per isakson
on 23 Apr 2014
Edited by per isakson
on 23 Apr 2014
 Accepted Answer

Problem: extracting first day of each month from one year of hourly data (two missing hours). One alternative is to compare day number
%%Create sample data
vec = datevec( '2010-01-01 00:00', 'yyyy-mm-dd HH:MM' );
num = datenum( vec(1),vec(2),vec(3),[vec(4)+transpose([0:8757])],0,0 );
str = datestr( num, 'yyyy-mm-dd HH:MM' );
cac = cat( 2, num2cell( str, 2 ), num2cell( ones(8758,7) ) );
%%find first day of each month
vec = datevec( cac(:,1), 'yyyy-mm-dd HH:MM' );
is1 = vec(:,3)==1;
ix1 = find( is1 );
%%extract rows
selection = cac( is1, : );
or comparing characters
str = char( cac(:,1) );
cel = num2cell( str( :, 9:10 ), 2 );
is1 = strcmp( '01', cel );
IMO: Neither of these two solutions is simple

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