for loop sequence from the matrix
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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 Kommentare
Mathieu NOE
am 22 Jan. 2021
hello
ok, but what are we suppose to do in the code , especially for i that is not present in the A vector ?
Turbulence Analysis
am 22 Jan. 2021
Masoud Dorvash
am 22 Jan. 2021
Bearbeitet: Masoud Dorvash
am 22 Jan. 2021
Hi there,
I really can't unerstand your question, you want that the for loop generates a new matrix which elements are from the A matrix with the position i ?
Turbulence Analysis
am 22 Jan. 2021
As an aside, you should replace all of that complex IF/ELSEIF/ELSE, NUM2STR, and STRCAT with this:
fname = sprintf('B%05d.vc7',j)
Turbulence Analysis
am 22 Jan. 2021
Masoud Dorvash
am 22 Jan. 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.
Turbulence Analysis
am 22 Jan. 2021
Antworten (1)
Sai Veeramachaneni
am 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
%Required tasks here.
end
end
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