How do I take the 'left' side of a matrix? (row-wise)

General question:

imagine my matrix being:

a =

1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6

then i want to get a matrix:

b =

1 2 3

1 2 3

1 2 3

More specifically:

I have a matrix of size n x 59577, a for loop adds a new row of data to the matrix depending how many audio files are being read.

numSamples = 59577; %the length of the audio files in samples
files = [1,3]; %These numbers can be chosen (values point to a different audio file)
nLoadedFiles = length(files);

for ii = 1:nLoadedFiles
    %Get the audio samples from the cell array and store them as rows in 'samples' 
    samples(ii,:) = audioData{files(ii)};
    %Take the fft of all loaded samples
    ffts(ii,:) = fft(samples(ii,:));
    SSBs(ii,:) = ffts(ii(1:numSamples/2),:); %This gives: Index exceeds the number of array elements (1).
    %So im looking for how to declare 
    SSBs = %SHOULD BE A COPY OF THE ROWS IN FFTS, BUT THEN ONLY THE FIRST HALF, IE. (1:numSamples/2)
end

Maybe the error is caused by numSamples being odd? (59577)

Answer 1

[KSSV]

For your first example:

b = a(:,1:3) ; 

If A is your matrix..if your first first n columns

B = A(:,1:n) ; 

If you want first n rows..

C = A(1:n,:) ; 

[Niels van Dalen]

I get this, I think I'm going wrong at the row indexing as

SSBs(ii,:) = ffts(ii(1:numSamples/2),:)

Or maybe at the fact that numSamples/2 = 29788.5 ? (which doesn't seem like a proper index XD)

[KSSV]

Yes theindex is a fraction..you will get error..use

SSBs{ii} = ffts(1:round(numSamples/2),:) ; 

[Niels van Dalen]

Thanks for the help :)