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How to access a element of a dynamically created array

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Akram RAYRI
Akram RAYRI 2022-4-12
评论： Akram RAYRI 2022-4-13
Hi all, I have created a set of arrays of dimension 1*3 using the following code :
BS = 3; area = 300;
X1 = area * rand(1,BS) -area/2 ;
Y1 = area * rand(1,BS) -area/2 ;
for i = 1:BS
eval(['BS' num2str(i) '= [X1(i) Y1(i) 1] '])
end
and i would like to dynamically access the third element of BS"i". Let's say that BS1(3) = 100, and BS2(3) = 150 and BS3(3) = -200. How I would realize that dynamically by using a for loop ?
  1 个评论
DGM
DGM 2022-4-12
编辑：DGM 2022-4-12
You know what's better than solving an unnecessary problem?
Not creating the problem in the first place.
https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval
Side note: There's not much point accessing BSx(3). You explicitly set them all to 1.

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DGM
DGM 2022-4-12
Ran in:
Just use an array
BS = 3;
area = 300;
BSx = [area*(rand(BS,2) - 0.5) ones(BS,1)]
BSx = 3×3
49.0217 81.9383 1.0000 -61.5512 72.0274 1.0000 -98.1272 -102.9906 1.0000
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DGM
DGM 2022-4-13
Ran in:
I never said it couldn't be done; I said it was unnecessary. You're defeating the most basic functionality of the language by embedding indexing information into the variable name.
Accessing the row vectors within a matrix is easier, faster (by about three orders of magnitude), and more flexible than doing the same with a bunch of dynamically accessed named vectors.
BS = 3;
area = 300;
BSx = [area*(rand(BS,2) - 0.5) ones(BS,1)]
BSx = 3×3
58.9408 15.3944 1.0000 -64.1582 65.8407 1.0000 -81.9446 -23.0681 1.0000
You want the first element of the second vector?
BSx(2,1)
ans = -64.1582
Want the third vector in whole?
BSx(3,:)
ans = 1×3
-81.9446 -23.0681 1.0000
Want all x values?
BSx(:,1)
ans = 3×1
58.9408 -64.1582 -81.9446
It's your choice.
Akram RAYRI
Akram RAYRI 2022-4-13
Yes I saw what you meant. Thank you so much for your reply anyways ;)

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