Beginner's roblems with arrays

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.

Answer 1

[Sean de Wolski]

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

[Matt Fig]

It is only by coincidence that V(2)==THETA(2) that this works.

[Morten Jensen]

hmm, seems like I most have missed some. A copy/paste of your reply works.

thanks.

[Morten Jensen]

@ Matt; V(2) and THETA(2) won't neccessarily always be the same. but the sizes of them will.

[Matt Fig]

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]

I see you resolved this below...

[Morten Jensen]

It was my mistake..

It is supposed to be like this:

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)) = THETA(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);

[Matt Fig]

See my comments below about the inefficient rewriting going on in this solution.

Answer 2

[Matt Fig]

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

[Morten Jensen]

Yes, a mistake from my side. and yes, there will always be 3 rows. only length can change.

[Matt Fig]

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...