for loop sequence from the matrix
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Turbulence Analysis
el 22 de En. de 2021
Respondida: Sai Veeramachaneni
el 27 de En. de 2021
Hi,
I intend to execute the for loop, in which sequence for given index needs to be called from a matrix. For e.g. I have a matrix named 'A', it got the discontinous numbers, like 10 to 19, then 120 to 150 and 238 to 247 so on.. now I ned to exeute the for loop with +5 and - 5 from the point where discontinuity exists, e.g. as shown in below code
I have attached the matrix A here. Please help with this...
for i = [5:1:15 115:1:125 233:1:243]
% my code
end
8 comentarios
Masoud Dorvash
el 22 de En. de 2021
I'm not sure if this example can help you or not.
firstSeq = 5:10;
secondSeq = 50:55;
thirdSeq = 500:505;
seq = [firstSeq secondSeq thirdSeq];
B = seq;
for i = 1:length(firstSeq)
B(i) = seq(i) - 5;
end
for j = length(firstSeq)+1:length(firstSeq)+length(secondSeq)
B(j) =seq(j) - 10;
end
for k = length(firstSeq)+length(secondSeq)+1:length(firstSeq)+length(secondSeq)+length(thirdSeq)
B(k) =seq(k) - 100;
end
but what you asked is somehow needs an if in the loop because the first sequence of the Amatrix can get 10 numbers but as you said you want the second one to get 9 numbers so you need to add a simple if in the for loop.
This is what you get in the result,
seq = 5 6 7 8 9 10 50 51 52 53 54 55 500 501 502 503 504 505
B = 0 1 2 3 4 5 40 41 42 43 44 45 400 401 402 403 404 405
Hope this works for you.
Respuestas (1)
Sai Veeramachaneni
el 27 de En. de 2021
Your question can be split into two parts.
- Identifying discontinous numbers in the matrix A.
- Exeute the for loop with +5 and - 5 from each discontinuous number.
Step1:
discontinous = [A(1)]%Stores the vector of discontinous numbers
for i = 2:numel(A)
discontinous = [discontinous A(i)];
end
Step2:
for i = 1:numel(discontinous)
for j = discontinous(i)-5:discontinous(i)+5
%Required tasks here.
end
end
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