Help with looping problem

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Tom
Tom el 23 de Oct. de 2012
Here I am defining x and y coordinates of rays vertically approaching a boundary (currently sin(x))
x = 0:0.1:pi
rx1 = [x(1) x(1)] % x-coordinates, incoming ray 1
ry1 = [y0 sin(x(1))] % y-coordinates, incoming ray 1
rx2 = [x(2) x(2)] % x-coordinates, incoming ray 2
ry2 = [y0 sin(x(2))] % y-coordinates, incoming ray 2
rx3 = [x(3) x(3)] % x-coordinates, incoming ray 3
ry3 = [y0 sin(x(3))] % y-coordinates, incoming ray 3
and so on for the length of vector (x) = 32
Clearly this isn't a good way to write the program. The alternative approach I have tried to use is the following:
i = 1
for i = 1:length(x)
rx(i) = [x(i) x(i)]
ry(i) = [y0 sin(x(i))]
end
Which returns the error:
In an assignment A(I) = B, the number of elements in B and I must be the same.
So clearly this isn't the way to achieve what I was trying to do. I would greatly appreciate if someone could help me resolve this, as I have been stuck on it for some time.

Respuesta aceptada

Matt Fig
Matt Fig el 23 de Oct. de 2012
Why would you want to create 64 variables like that? What a mess it would be to deal with in further steps. Perhaps two cell arrays could get the job done:
for ii = 1:length(x)
rx{ii} = [x(ii) x(ii)]; % rx is a cell array.
ry{ii} = [y0 sin(x(ii))];
end
Now the really important question is why do you need to do this at all? What is your next step that requires you to have these arrays as individuals, rather than just using x and sin(x)?
  6 comentarios
Tom
Tom el 23 de Oct. de 2012
I haven't encountered this 'line' parameter yet. How come it is plotted without even being asked?
Matt Fig
Matt Fig el 23 de Oct. de 2012
LINE is not a parameter, it is a function.
help line

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Más respuestas (3)

Matt Kindig
Matt Kindig el 23 de Oct. de 2012
Editada: Matt Kindig el 23 de Oct. de 2012
I'm not entirely clear what you want, but I think this will do the trick:
x = (0:0.1:pi)';
rx = [x, x];
ry = [ repmat(y0, size(x)), sin(x)];
This implements the vector as specified in the for loop you have shown, which is the preferred way to specify matrix data in Matlab (not creating variables for each line, as your rx1, rx2, etc. example shows).
  1 comentario
Tom
Tom el 23 de Oct. de 2012
Editada: Walter Roberson el 23 de Oct. de 2012
Thanks for the swift response!
That didn't seem to work, perhaps it would make my problem clearer if you run the following:
x = 0:0.1:pi;
y = sin(x);
plot(x,y,'k-')
hold on
rx1 = [x(1) x(1)]
ry1 = [-1 sin(x(1))]
rx2 = [x(2) x(2)]
ry2 = [-1 sin(x(2))]
rx3 = [x(3) x(3)]
ry3 = [-1 sin(x(3))]
plot(rx1,ry1,rx2,ry2,rx3,ry3)
This is what I am trying to achieve (although I have only shown the first 3 vertical lines).
My inexperienced attempt to achieve this with a For loop goes:
for i = 1:length(x)
rx(i) = [x(i) x(i)]
ry(i) = [y0 sin(x(i))]
end
Which returns the error shown above, making clear I'm going completely wrong. Any idea what I should be doing?

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Walter Roberson
Walter Roberson el 23 de Oct. de 2012

Azzi Abdelmalek
Azzi Abdelmalek el 23 de Oct. de 2012
y0=1
x = 0:0.1:pi
n=length(x);
eval(sprintf('rx%d=[x(%d) x(%d)],',repmat((1:n),3,1)))
eval(sprintf('ry%d=[y0 sin(x(%d))],',repmat((1:n),2,1)))

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