An easy way to modify matrix elements?

I've created a matrix, P, representing products (the rows set by NCE = 1:NCEMAX) over time (20 years - columns). I launch 1 product/year, some of which will fails 4 years into production (randomly 3x1 matrix). It might look like this without failure (a 1 represents in manufacture)

NCEMAX = 3;
P=[...
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
Failure = [1;0;1]

The matrix I want to create would be, as the second product fails 4 years after it starts (ie year 5 as it is NCE2)

NCE2 = [...
    1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
    0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]

I've tried

P(: , NCE+4:end) = P(: , NCE+4:end) .* Failure;

but the NCE+4 doesn't shift up each row, i.e. NCE+4 always = 5 not 5,6 then 7.

What am I doing wrong? Any help much appreciated.

[Torsten]

Please include your executable code.

[Stephen23]

NCE is a vector. Adding 4 to NCE returns another vector. Using a vector as any input to the colon operator ignores all elements of the vector except the first element, as documented here:

https://www.mathworks.com/help/matlab/ref/colon.html#description

" If you specify nonscalar arrays, then MATLAB interprets j:i:k as j(1):i(1):k(1)."

That explains the behavior you observe. It is not clear from your question what you expect to happen.

[Image Analyst]

What does Failure represent? Like if it is 1 then that row fails for years 4 and later so P for that row would be 1 for columns 1,2 and 3, then 1 for 4-20?

What exactly is supposed to be shifting upwards? Some rows get shifted up, like row 3 moves into row 2 or something?????

So P=1 for no failure and P=0 for failure? And what value indicates failure in the failure vector 1 or 0?

Why does the first column and second of NCE2 have some zeros in some rows before the 1's start? Is that because the product has not yet been produced for those years?

[Andrew Rutter]

Yes sorry, it is hard to write clearly, Failure means product fails in development, typically after 4 years, so in my example the first and third make it into production and the second fails, hence it is made for 4 years then has zero production, In P, 1 represents it being made and 0 a fail.

As one product starts development each year, this produces a zero in column 1 and 2 for second and third products as they havent started development.

hope this makes sense.

Answer 1

[Image Analyst]

Perhaps this:

NCEMAX = 3;
P = ones(NCEMAX, 20);
Failure = [1;0;1];
[rows, columns] = size(P);
NCE2 = zeros(rows, columns);
for row = 1 : rows % For each year.
    NCE2(row, row:end) = 1; % Initialize to no failure
    if Failure(row) == 0
        % Failed at year 4 after introduction
        NCE2(row, row+4:end) = 0;
    end
end
NCE2 % Show in command window.

NCE2 = 3×20

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

[Andrew Rutter]

This looks promising, I will give a try and get back to you...thank you for helping

[Andrew Rutter]

Its works - thank you so much....best regards, Andrew

Answer 2

[Torsten]

for NCE = 1:NCEMAX
  if Failure(NCE) == 0
    P(NCE,NCE+4:end) = 0
  end if
end