I have tried to make a loop where it displays all the answers in a matrix but it doesn't give me the 21x3 matrix. Am i using wrong commands or am i missing something.
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merlin kelly
el 14 de Nov. de 2018
Comentada: Luna
el 15 de Nov. de 2018
t=0;
for v=[t,sin(t),cos(t)]
if t<21
disp(v)
t=t+1;
end
end
2 comentarios
Respuesta aceptada
Luna
el 14 de Nov. de 2018
Hi Merlin,
try this code below:
v = [];
for t = 0:1:20
v(t+1,:) = [t,sin(t),cos(t)] ;
end
2 comentarios
Luna
el 15 de Nov. de 2018
actually I prefer to define v = nan(21,3). Pre-allocation of the size increases the performance, and speed up the for loop.
defining empty ( v = [] ) is something about usage.
For example: if you use your for loop in if-else block, defining empty in the beginning helps you understand whether it goes to if or else block, etc.
You can easily check if isempty(v) that your matrix filled with values or remained empty at end of the process.
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