How to collect individual bunches of data in a column of a matrix

Hi,

I have a matrix where I would like to collect the first 203 rows of a column into a variable, say e1, then skip the next 813 rows, then collect and add the next 203 rows into e1 again, then skip the next 813 rows... and so on until the end of the column.

I am new to MatLab and could not figure this out myself. Could someone suggest how to do this please?

Thanks

[Adam]

idx = repmat( [ones(203,1); zeros(813,1)], n, 1 );
newColumn = oldcolumn( idx );

Something like that would work, where n is however many times you need to repeat the sequence, such that it still fits within oldColumn.

Answer 1

[Guillaume]

Assuming that the length of your column vector is a multiple of 203+813 (if not, what should be done for the last bit that is too short?), then a possible way is to reshape your vector into columns of 203+813 elements and only keep the first 203 rows before reshaping back into a column:

desired = reshape(yourcolumnvector, 203+813, []);
desired = reshape(desired(1:203, :), [], 1)

Another is to simply build the whole list of indices:

indices = (1:203)' + (0:numel(yourcolumnvector)/(203+813)-1) * (203+813);
desired = yourcolumnvector(indices)

[Darren Nutting]

Hi,

My data is not divisible by 203+813 so I have been trying the second method you proposed. I am having trouble telling if it is working as my data set contains around 250k points.

Could you please break down your logic in the second method so I can understand what you have written better?

Thank you

[Guillaume]

My data is not divisible by 203+813

Then what should be done for the last bit of data? If mod(numel(yourvector), 203+813) is >= 203, then it means that you don't have enough skip element, which I guess is not too much of a problem, but if it's < 203, then you don't have enough elements to put in your last column.

The second method simply creates a column vector of indices 1:203, and a row vector of skip 0:(203+813):end. Adding the two together gives you a 2D matrix of indices to extract (because of implicit expansion).

[Darren Nutting]

Thanks :) And I have managed to use your second method to get it to work now :)