i have 3x920 double that call A
for k=1:length(A)-1 %% same as 1:919
Ax = A(2, k+1) - A(2,k)
end
Why is there only one value in the workspace?
i want something that 1x919 double
ex) Ax | 1x919

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Turlough Hughes
Turlough Hughes 2021년 8월 29일
편집: Turlough Hughes 2021년 8월 29일

0 개 추천

There's only one value because your code is working as follows:
for ii = 1:3
A = ii + 1
end
A = 2
A = 3
A = 4
Matlab is not told where to store the data inside A, so A(1) just gets changed each time the loop iterates. You need to use the loop variable, ii, to indicate the index position in A where you would like to store data for the current iteration.
Let's try this again:
for ii = 1:3
A(ii) = ii + 1
end
A = 2
A = 1×2
2 3
A = 1×3
2 3 4
We can do better though; it's important to preallocate the space to your variable when it's possible to do so:
A = zeros(1,3)
A = 1×3
0 0 0
for ii = 1:3
A(ii) = ii + 1
end
A = 1×3
2 0 0
A = 1×3
2 3 0
A = 1×3
2 3 4

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