Multiple outputs from a for loop

2022 6 21
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更新时间：2022 6 21
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Basic problem, but I can't find the solution:
consider basic a for loop:
for i = 1:5
z(i)=3*i
end
It produces a vector z= [3 6 9 12 15]. What I need is all the components of the vector z as a sperate variables(scalars), let,s say
z1=3, z2=6 and so on. How can I do that?
Thanks in advance
Roger Kalvig

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Stephen23
Stephen23 2022-6-21
"What I need is all the components of the vector z as a sperate variables(scalars)"
Why?
What do you imagine that you can do with lots of scalars, that is not possible using indexing?
"say z1=3, z2=6 and so on."
https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval
KSSV
KSSV 2022-6-21
Already you have that in hand.
z(1), z(2),...z(n) etc.,
Roger Kalvig
Roger Kalvig 2022-6-21
Ok seems I need to describe my entire problem instead of decomposing it into pieces.
The goal is to perform numerical integracion of two functions. One of them is my function fcn looking like this:
function y = fcn(x)
Bext = 0.6;
G = 10.553;
if x < 0
y = 0;
elseif x == 0
y = 100;
elseif x == 180
y = 100;
elseif x > 180
y = 0;
else
y = 1./(G.*Bext.*(sind(x)));
end
end
which is simple relationship defined for 3 ranges and 2 x points.
The second function is Gaussian function:
function y = myGausstest(x)
A = 10;
w = 20;
xc = 90;
y = A.*exp(-0.5*((x-xc)./w).^2);
end
Now fcn is always constant, but Gauss is moving along x axis. All parameters of Gauss function remain constat except xc. So my idea was to create a for loop:
for xc = -100:10:280 (some range suiting fcn function)
"do integration here of the product of both functions fcn*myGausstest for integral limts (xc-d,xc+d)" using trapz or integral
end
Finally I need do plot the result
plot(x, result of integration)
Any help is appreciated.
Roger Kalvig

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Stephen23
Stephen23 2022-6-21
编辑：Stephen23 2022-6-21
Ran in:

0 个投票

Ran in:
Rather than distracting with anti-pattern numbered variables (like you asked about), instead you should just use basic, normal, efficient, easy, simple, standard indexing. Indexing is how MATLAB works best.
d = 5;
xc = -100:10:280; % (some range suiting fcn function)
iv = nan(1,numel(xc));
fh = @(x) fcn(x) .* myGausstest(x);
for k = 1:numel(xc)
% do integration here of the product of both functions fcn*myGausstest for integral limts (xc-d,xc+d)" using trapz or integral
iv(k) = integral(fh,xc(k)-d,xc(k)+d);
end
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 6.7e-03. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
Warning: Minimum step size reached near x = 180. There may be a singularity, or the tolerances may be too tight for this problem.
plot(xc,iv)
function y = fcn(x)
Bext = 0.6;
G = 10.553;
y = 1./(G.*Bext.*sind(x));
y(x<0) = 0;
y(x==0) = 100;
y(x==180) = 100;
y(x>180) = 0;
end
function y = myGausstest(x)
A = 10;
w = 20;
xc = 90;
y = A.*exp(-0.5*((x-xc)./w).^2);
end

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Roger Kalvig
Roger Kalvig 2022-6-21
One problem i can see is I already defined xc within
function y = myGausstest(x)
A = 10;
w = 20;
xc = 90;
y = A.*exp(-0.5*((x-xc)./w).^2);
end
so idea is to express Gauss in fh explicitely:
fh = @(x) fcn(x) .* A.*exp(-0.5*((x-xc)./w).^2);
but again, matrix dimensions are incorrect.
Stephen23
Stephen23 2022-6-21
编辑：Stephen23 2022-6-21
Ran in:
Ran in:
"One problem i can see is I already defined xc within"
Presumably you only intend the xc inside the function to refer to a single value, not the entire vector, in which case you can easily pass the value by parameterizing the function:
https://www.mathworks.com/help/matlab/math/parameterizing-functions.html
For example:
d = 5;
xc = -100:10:280; % (some range suiting fcn function)
iv = nan(1,numel(xc));
for k = 1:numel(xc)
fh = @(x) fcn(x) .* myGausstest(x,xc(k)); % parameterize function
iv(k) = integral(fh,xc(k)-d,xc(k)+d);
end
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 1.7e+02. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
Warning: Minimum step size reached near x = 180. There may be a singularity, or the tolerances may be too tight for this problem.
plot(xc,iv)
function y = fcn(x)
Bext = 0.6;
G = 10.553;
y = 1./(G.*Bext.*sind(x));
y(x<0) = 0;
y(x==0) = 100;
y(x==180) = 100;
y(x>180) = 0;
end
function y = myGausstest(x,xc)
A = 10;
w = 20;
y = A.*exp(-0.5*((x-xc)./w).^2);
end
Roger Kalvig
Roger Kalvig 2022-6-21
Thanks very much for help

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