Hi, I need to do something like this:
a=[1 2 3 4 5 6 7 8 9 10];
for i=1:lenght(a)
for ii=i-2:i+2
x=a(ii);
end
end
I found a solution doing something like this:
a=[1 2 3 4 5 6 7 8 9 10];
for i=1:lenght(a)
if i==1
elseif i==2
elseif i=>3&&i<length(a)-2
elseif i==9
elseif i==10
end
end
I need to do the iteration only in the range +- 2 of i.The problem is that a(-1) ,a(0),a(11) and a(12) doesn't exist . . Is there any simple way to solve that problem without using the if conditions?

2 Comments

qué quieres hacer específicamente?
what do you want to do specifically?
Estoy tratando de hacer un código para encontrar la cantidad de barras de acero suficientes para cubrir la demanda de acero de un elemento de concreto reforzado. El código combina los tipos de barras que existen y me dice cuántas de cada una son necesarias para lograr el objetivo. Adicionalmente, la combinación de barras se hace solamente con aquellas que están dos posiciones por encima o por debajo. Por ejemplo:
Se tienen barras No.2, No.3,No.4... hasta No.18
Si quiero combinar barras No.4 solo puedo hacerlo con barras No.2,No.3,No.4,No.5 y No.6. El problema está cuando debo combinar las barras No.2 ya que en este caso solo puedo hacerlo con No.2, No.3 y No.4.

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 Accepted Answer

a = 1:10;
n = numel(a);
for kk = 1:n
for ii = max(1,kk-2):min(n,kk+2)
x = a(ii);
end
end

1 Comment

Thanks a lot, that is what I was looking for

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More Answers (1)

How about limiting the loop from 3 to length(a)-2, like the following?
Also, if your calculation process in the inner loop can be re-written without using for-loop, I would strongly recommend doing so.
for k = 3:lenght(a)-2
for kk = k-2:k+2
% Your calculation process
end
end

1 Comment

This is a partial solution to the problem but in that case i won't have the first two iterations and the last two ones. For the first iteration the internal for loop must be:
for k=1:lenght(a)
for kk=k:k+2
end
The second iteration:
for k=1:lenght(a)
for kk=k-1:k+2
end
The same logic applies to the last two iterations

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