Help with looping problem

Tom
2012 10 23
4 个回答
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0 个投票

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.

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Matt Fig
Matt Fig 2012-10-23

0 个投票

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)?

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Tom
Tom 2012-10-23
编辑：Tom 2012-10-23
Haha! I wouldn't be surprised to hear I have gone the complete wrong way about this. Perhaps I could ask you then: How would you plot a number of vertical lines in increments along the x-axis, which each terminate when they reach the function sin(x)?
Really appreciate the help by the way.
p.s. when I ran your suggestion I received the error: Error using plot Not enough input arguments.
Matt Fig
Matt Fig 2012-10-23
Do this instead:
x = 0:0.1:pi;
y = sin(x);
plot(x,y,'k-')
hold on
for ii = 1:length(x)
line([x(ii) x(ii)],[-1 y(ii)])
end
Tom
Tom 2012-10-23
That is exactly what I needed. I cannot thank you enough!
Matt Fig
Matt Fig 2012-10-23
编辑：Matt Fig 2012-10-23
Or without the FOR loop at all:
x = 0:0.1:pi;
y = sin(x);
plot(x,y,'k-')
L = line([x;x],[-ones(1,length(x));y]);
set(L,'color','b')
Tom
Tom 2012-10-23
I haven't encountered this 'line' parameter yet. How come it is plotted without even being asked?
Matt Fig
Matt Fig 2012-10-23
LINE is not a parameter, it is a function.
help line

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更多回答（3 个）

Matt Kindig
Matt Kindig 2012-10-23
编辑：Matt Kindig 2012-10-23

0 个投票

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).

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Tom
Tom 2012-10-23
编辑：Walter Roberson 2012-10-23
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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Azzi Abdelmalek
Azzi Abdelmalek 2012-10-23

0 个投票

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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