Help with a loop

1 vue (au cours des 30 derniers jours)
Konstantina Vaitsi
Konstantina Vaitsi le 13 Mai 2020
I need to write a script where I have two matrixes. One S = [12 4 6 4;16 12 2 3;0 4 10 12;4 12 21 3] and one CLN = 50*ones(4). I need to create a loop where in every step all the cells in S matrix are decreased by 2 and a new CLN matrix is calculated depending on the S and the previous CLN. Specifically,
CLN(i,j) = CLN(i,j) -0.5*S(i,j) if S(i,j) <5,
CLN(i,j) = CLN(i,j)+ 0.5*S(i,j) if S(i,j) > =5
and CLN(i,j) = 0 if CLN(i,j)-0.5*S(i,j)<0.
This is what I have written so far but it doesn't work
S = [12 4 6 4;16 12 2 3;0 4 10 12;4 12 21 3];
CLN = 50*ones(4);
i = 1:4;
j = 1:4;
for t = 0:4
B = [S - 2*t];
B(B<0)=0
Y = zeros(size(B));
Y(B<5) = CLN - 0.5*B(B<5);
Y(B>=5) = CLN + 0.5*B(B>=5);
Y(CLN - 0.5*B<0) = 0(CLN - 0.5*B<0)
endfor

Réponse acceptée

William Alberg
William Alberg le 15 Mai 2020
S = [12 4 6 4;16 12 2 3;0 4 10 12;4 12 21 3];
CLN = 50*ones(4);
for i = 1:4
S = S - 2;
CNL_new = CLN;
% If S is below 5
index = S < 5;
CNL_new(index) = CLN(index) - 0.5*S(index);
% If S is above or equal to 5
index = S >= 5;
CNL_new(index) = CLN(index) + 0.5*S(index);
% if CLN - 0.5*S < 0
index = (CLN - 0.5*S) < 0;
CNL_new(index) = 0;
CLN = CNL_new;
end
Does this help?
  1 commentaire
Konstantina Vaitsi
Konstantina Vaitsi le 15 Mai 2020
I spent days on it not knowning what to do. I made a few adjustments and it finally works! Thanks a million :)

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by