# for loop sequence from the matrix

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Turbulence Analysis on 22 Jan 2021
Answered: Sai Veeramachaneni on 27 Jan 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
for i = [5:1:15 115:1:125 233:1:243]
% my code
end
Masoud Dorvash on 22 Jan 2021
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.
Turbulence Analysis on 22 Jan 2021
Sorry, this is different.
I got 10000 files inside the folder, I need to read only selected files as per the numbers prsent in the matirx A. If you open matrix, A it starts from 10 to 19, then 120 to 150 and then 238 to 247 with the last sequence being 9914 to 9953. Here, i have to read the files based discontinuity, for e.g. it starts with 10, so first seqeunce for loop should be 5:15, after 19 its starts with 120 to 150 and agian discontinous. So my second sequence would be 115: 125. Basically I need to read five files before and after the discontinous number...

Sai Veeramachaneni on 27 Jan 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
end
end