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

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!

Answer 1

[per isakson]

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