Beginner's roblems with arrays
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Hi!
I asked a question here a few days ago and was a bit too hasty in accepting the answer since it only solved a part of my problem.
Basically, I need to combine some values from different arrays:
K = 130;
V = [0.9 0 .5];
THETA = [pi/4 0 3*pi/4];
T = [0 33 57];
So that I have a Vnew and THETAnew that are 1xK and whose values change at T(1:end). So far I've managed to do it crudely like this:
timestep= ones(3,K);
timestep(1,:) = 0:K-1;
timestep(2,1:T(1,2) = V(1,1);
timestep(2,T(1,2)+1:T(1,3) = V(1,2);
timestep(2,T(1,3)+1:end = V(1,3);
timestep(3,1:T(1,2) = THETA(1,1);
timestep(3,T(1,2)+1:T(1,3) = V(1,2);
timestep(3,T(1,3)+1:end = THETA(1,3);
It works, however I'd like to make the code more neat and able to to handle any lengths of V, THETA and T (length(V) = length(THETA) = length(T) will always be true). For this, I've tried the following:
timestep = zeros(3,K);
timestep(1,:) = 0:K-1;
for l = 1:size(T,2)
timestep(2,T(1,l)+1:end) = V(l);
timestep(3,T(1,l)+1:end) = THETA(l);
end
But I cannot get this piece of code to work. MATLAB simply returns an error message when I try to run it.
Any help would be very appreciated.
回答（2 个）
Sean de Wolski
2011-6-23
I'm not getting any errors after I add a few parenthesis:
K = 130;
V = [0.9 0 .5];
THETA = [pi/4 0 3*pi/4];
T = [0 33 57];
timestep= ones(3,K);
timestep(1,:) = 0:K-1;
timestep(2,1:T(1,2)) = V(1,1);
timestep(2,T(1,2)+1:T(1,3)) = V(1,2);
timestep(2,T(1,3)+1:end) = V(1,3);
timestep(3,1:T(1,2)) = THETA(1,1);
timestep(3,T(1,2)+1:T(1,3)) = V(1,2);
timestep(3,T(1,3)+1:end) = THETA(1,3);
timestep2 = zeros(3,K);
timestep2(1,:) = 0:K-1;
for l = 1:size(T,2)
timestep2(2,T(1,l)+1:end) = V(l);
timestep2(3,T(1,l)+1:end) = THETA(l);
end
%%Check
isequal(timestep,timestep2)
%ans = 1
7 个评论
Matt Fig
2011-6-23
It is only by coincidence that V(2)==THETA(2) that this works.
Morten Jensen
2011-6-23
Morten Jensen
2011-6-23
Matt Fig
2011-6-23
So then the loops won't work. This is because, as I state below, you are assigning:
timestep(3,T(1,2)+1:T(1,3)) = V(1,2);
instead of:
timestep(3,T(1,2)+1:T(1,3)) = THETA(1,2);
If there are more occurrences of this type, looping will be very difficult.
Matt Fig
2011-6-23
I see you resolved this below...
Morten Jensen
2011-6-23
Matt Fig
2011-6-23
See my comments below about the inefficient rewriting going on in this solution.
Matt Fig
2011-6-23
I cannot see a clear pattern since you are setting
timestep(3,T(1,2)+1:T(1,3)) = V(1,2);
Now if it was THETA(2), that we could work with. Will there always be three rows?
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EDIT With clarifications in comments...
Here is a way to do it without re-writing parts of the vector over and over. This re-writing could be very inefficient for large arrays. This does assume T(1) is 0.
timestep2 = ones(3,K);
timestep2(1,:) = 0:K-1;
for ii = 1:length(V)-1
timestep2(2,T(ii)+1:T(ii+1)) = V(ii);
timestep2(3,T(ii)+1:T(ii+1)) = THETA(ii);
end
timestep2(2,T(ii+1)+1:K) = V(ii+1);
timestep2(3,T(ii+1)+1:K) = THETA(ii+1);
%
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EDIT To illustrate the inefficiency discussed above:
K = 130000;
V = rand(1,5000);
THETA = rand(1,5000);
T = [0 sort(round(rand(1,4999)*13000))];
tic
timestep3 = zeros(3,K);
timestep3(1,:) = 0:K-1;
for l = 1:size(T,2)
timestep3(2,T(1,l)+1:end) = V(l);
timestep3(3,T(1,l)+1:end) = THETA(l);
end
toc
tic
timestep2 = ones(3,K);
timestep2(1,:) = 0:K-1;
for ii = 1:length(V)-1
timestep2(2,T(ii)+1:T(ii+1)) = V(ii);
timestep2(3,T(ii)+1:T(ii+1)) = THETA(ii);
end
timestep2(2,T(ii+1)+1:K) = V(ii+1);
timestep2(3,T(ii+1)+1:K) = THETA(ii+1);
toc
isequal(timestep2,timestep3)
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I get:
Elapsed time is 4.148093 seconds. % re-writing
Elapsed time is 0.014087 seconds. % not re-writing
2 个评论
Morten Jensen
2011-6-23
Matt Fig
2011-6-23
Ah, I see... Now it makes sense. Please be careful when posting questions in the future. Besides this mistake, you missed a bunch of parenthesis as well...
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2011-6-23
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