A shortcut for recurring if statements

Hello, I have a case where I don't know how many elements a vector will have and then depending on different size vectors I'll either include something or not include something.

For example, what I have now is just

    if numtable2rows >=2
        tablerow(2,:) = tablerow(1,:) + [0 50 0 0];
    end
    
    if  numtable2rows >=3
        tablerow(3,:) = tablerow(2,:) + [0 50 0 0];
    end
    
    if numtable2rows >=4
        tablerow(4,:) = tablerow(3,:) + [0 50 0 0];
    end
    
    if numtable2rows >=5
        tablerow(5,:) = tablerow(4,:) + [0 50 0 0];
    end

and I'm just expecting no more than 5 elements in numtable2row.

Does anyone know a more efficient way to do this?

Thank you in advance!

[Stephen23]

@tiwwexx: that algorithm seems unusual, because it discards the existing row data. Is that correct? In any case, please provide some example data, with both input and output matrices.

[tiwwexx]

More like builds off the original row data. Some example data would be "in one case I have 3 rows so numtablerows = 3". I also have a tablerow(1,:) and know that the height of each row is 50 pixels. Hence, if in this table I'm using ocr on each row individually I'll just build all the other row areas from the one I already have.

Answer 1

[tiwwexx]

tablerow(1,:) = [a b c d];
for k= 1:numtable2rows
    tablerow(k+1,:) = tablerow(k,:) + [0 50 0 0];
end

This works. Inspired by Walter Roberson.

[Walter Roberson]

That is not going to work for multiple reasons. size() returns a vector. tablerow(k) is a scalar. You start from 1 instead of starting from the current size and going to 5

[tiwwexx]

There we go. I was sloppy in writing this down. numtbale2rows is a number, say '3' one time and '5' another time. then I already have a tablerow(1,:) so I can build off that.

[Walter Roberson]

You fixed the size() but not the fact that tablerow(k) is a scalar.

tablerow(1,:) = [a b c d];
for k= 1:numtable2rows
    tablerow(k+1,:) = tablerow(k, :) + [0 50 0 0];
end

Answer 2

[Walter Roberson]

for K = size(tablerow,2)+1 : 5
   tablerow(K,:) = tablerow(K-1,:) + [0 50 0 0];
end