How to re-arrange swapped elements in an estimated vector as compared to a reference vector

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I have a reference vector u=[0.5 1 1.5 0.6981 1.3962 1.5707].
I estimate several vectors of the same size using a metaheuristic algorithm. But the problem is that sometimes the algorithm gives me vectors in which the positions of elements are swapped and sometimes there is no change in the positions i.e. sometimes positios of 1st and 2nd elements are interchanged, sometimes positions of 1st and 3rd elemtns are interchnaged and sometimes positions of 2nd and 3rd eleents are interchanged with each other in the estimated vectors.And if the changes ocuurs in the positions of two elements in the 1st three elements, the same change occurs in the last three elements also.But sometimes the positions are not interchnaged in the estimated vectors. I want to re-arrange the estimated vector elements if they interchange their positions. What check should I keep on the positions of the elements of the estimated vectors. Regards,
  3 Comments
Sadiq Akbar
Sadiq Akbar on 13 Nov 2019
Edited: the cyclist on 13 Nov 2019
I use Flower Pollination Algorithm. My work is to estimate my reference vector.
u=[0.5 1 1.5 0.6981 1.3962 1.5707]
When I run the algorithm, it gives me an estimated vector like
Est1=[0.499 1.002 1.5001 0.6890 1.3880 1.49998]
In this there is no change in the position of any element. But when I run it again, It gives me
Est2=[0.9999 0.5001 1.49999 1.3990 0.6999 1.5880]
As you can see here the position of 1st and 2nd elements are interchanged in Est2 as compared to u. Likewise when I run it again, it gives me
Est3=[0.5001 1.4981 1.0012 0.70001 1.57160 1.39990]
Here the positions of 2nd and 3rd elemnts are interchanged and so on. As is clear when there is no change of positions in two of the the 1st three elements, then no change occurs in the positions of the last three elements also as is clear in Est1. But when the 1st and 2nd elements interchange their positions in the 1st three elements, then the same happens pattern is followed in the last three elements also as in Est2. And the same can be observed in Est3. I want to rearrange these positions on the pattern of my vector u even if the algorithm gives me Est vector having changed positions.
Regards,

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Accepted Answer

the cyclist
the cyclist on 16 Nov 2019
Here is a different approach. It relies on the fact that you want the first three inputs to be in ascending numeric order, and ignores whether they are close to [0.5, 1.0, 1.5].
% Define indices for convenience
half_1st = 1:3;
half_2nd = 4:6;
% Reference vector
u=[0.5 1 1.5 0.6981 1.3962 1.5707];
% Some of the inputs
temp = [ ...
1.4293 1.6429 0.1220 1.3833 1.5643 0.7691
1.0385 0.5639 0.5502 1.5833 1.4221 1.5648
0.6842 1.4170 0.1005 1.3988 1.5756 0.0398
0.0156 1.3050 0.9022 0 1.5700 1.3977
0.7765 0.0962 1.3453 1.4011 0.7075 1.5688
1.2783 0.0559 1.1249 1.5817 0.4605 1.3814];
% Loop over the sample inputs
for nn = 1:6
[sortTemp, sortIndex] = sort(temp(nn,half_1st));
two(nn,:) = temp(nn,[half_1st(sortIndex) half_2nd(sortIndex)]);
end
I hope this one works for you.
  1 Comment
Sadiq Akbar
Sadiq Akbar on 16 Nov 2019
Thank you very much the cyclist. Indeed it worked. I am really very thankful to your heartedly help. Regards,

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More Answers (2)

the cyclist
the cyclist on 13 Nov 2019
Edited: the cyclist on 13 Nov 2019
% Inputs
u=[0.5 1 1.5 0.6981 1.3962 1.5707];
Est1=[0.499 1.002 1.5001 0.6890 1.3880 1.49998];
Est2=[0.9999 0.5001 1.49999 1.3990 0.6999 1.5880];
Est3=[0.5001 1.4981 1.0012 0.70001 1.57160 1.39990];
% Define a few parameters for convenience
half_1st = 1:3;
half_2nd = 4:6;
tol = 0.01;
% Find the elements of Est1, etc, that are near u
% EDITED for a mistake in the ordering of the ismembertol arguments
[tf1,idx1] = ismembertol(u(half_1st),Est1(half_1st),tol);
[tf2,idx2] = ismembertol(u(half_1st),Est2(half_1st),tol);
[tf3,idx3] = ismembertol(u(half_1st),Est3(half_1st),tol);
% Define the rearranged vectors
Est1_new = Est1([half_1st(idx1) half_2nd(idx1)]);
Est2_new = Est2([half_1st(idx2) half_2nd(idx2)]);
Est3_new = Est3([half_1st(idx3) half_2nd(idx3)]);
  11 Comments
the cyclist
the cyclist on 15 Nov 2019
Regarding my solution, you keep saying it does not work, and reporting the error that MATLAB is reporting. But have you tried to understand why you are getting that error, or how my solution should work in the first place?
I suggest you try adjusting the tolerance value, which might be too small. Maybe try setting
tol = 0.1; % Instead of tol = 0.01
and see if that solves it.

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Sadiq Akbar
Sadiq Akbar on 15 Nov 2019
Edited: the cyclist on 15 Nov 2019
half_1st = 1:3;
half_2nd = 4:6;
tol = 0.1;
u=[0.5 1 1.5 0.6981 1.3962 1.5707];
nn=0;
for n=100:199
nn=nn+1;
[out1,out2,out3]=flower(n,Remaining input arguments---)
temp(nn,:)=out1;
------------
-----------
%-------------Check for swapping-----------------
[tf,idx] = ismembertol(u(half_1st),temp(nn,half_1st),tol);
two(nn,:) = temp(nn,[half_1st(idx) half_2nd(idx)]);
end
When I run the above code, I get the following error even though I have reduced value of tol to 0.1 as per your instruction:
Subscript indices must either be real positive integers or logicals.
Error in servertestingMathworks (line 26)
two(nn,:) = temp(nn,[half_1st(idx) half_2nd(idx)]);
  3 Comments
the cyclist
the cyclist on 16 Nov 2019
OK, so it is clear (to me) why my algorithm isn't working. Do you see how your original example input
Est1=[0.499 1.002 1.5001 0.6890 1.3880 1.49998]
Est2=[0.9999 0.5001 1.49999 1.3990 0.6999 1.5880]
Est3=[0.5001 1.4981 1.0012 0.70001 1.57160 1.39990]
the values are all very close to [0.5 1.0 1.5]? The biggest discrepancy is only about 0.002! I assumed that that was always true, but your new inputs
1.4293 1.6429 0.1220 1.3833 1.5643 0.7691
1.0385 0.5639 0.5502 1.5833 1.4221 1.5648
0.6842 1.4170 0.1005 1.3988 1.5756 0.0398
0.0156 1.3050 0.9022 0 1.5700 1.3977
0.7765 0.0962 1.3453 1.4011 0.7075 1.5688
1.2783 0.0559 1.1249 1.5817 0.4605 1.3814
can be much further away from [0.5 1.0 1.5].
Neither my algorithm nor your algorithm, which both rely on proximity to the original vector, will work for this input. I have another idea that I will post later.

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