MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn moreOpportunities for recent engineering grads.

Apply Today**New to MATLAB?**

Asked by Usman Ali
on 7 May 2012

hello every one.. i am a bit new to matlab so have no vocabulary of commands or how thing work as vector matrices or etc etc.. so i would like to develop a prototype or an algorithm for scenario reduction in matlab( each row of a matrix is called as single scenario).

the steps of this algorithms are

1) for any given matrix, find the distance matrix which show euclidean distance between each and every row of original matrix.

2) find the min: value in distance matrix.

3) delete that row with min elemenet value and with min: probabilty of occurance.

4) construct seperate matrix which contain only these deleted rows

5) reconstruct distance matrix with preserved row (without deleted rows).

6) repeat untill the desired numb of rows are preserved.

So I started with a simple 3x3 matrix and i wrote few lines as follow.

--------------------------------------

% suppose any matrix e.g

x=[5 3 1; 2 5 6; 1 3 2];

dist_mat = squeeze(sqrt(sum(bsxfun(@minus,x,reshape(x',1,size(x,2),size(x,1))).^2,2)));

dist_mat(~dist_mat)=inf;

row_min = min(dist_mat);

min_value = min (row_min);

[r c]= find(d==min_value); % shows the position of the minimum element, r= which row and c= which column

so the results i get are

d =

Inf 6.1644 4.1231 6.1644 Inf 4.5826 4.1231 4.5826 Inf

min_value =

4.1231

r =

3 1

c =

1 3

thanks to 'Sean de Wolski' for correcting the dist_mat command.

in my e.g, the min value is in [3,1] and [1,3]. i.e D13=D31, this is correct.

let every row has its own probability, i.e P1= 0.2, P2=0.4, P3=0.4 (total P=1) since the min value = 4.1231 lying in 1st and 3rd row, we can delete either Row1 or Row3, but since the probabilty of Row1 < Row3 so row1 should be selected for deletion. i do not have any thing in mind how to delete this row, rearrange the dist_mat and write a new mat with this deleted row.

I m really sorry for so long post but i wanted to make thing more clear... i hope to get some +ve responses...

Thanks alot in Advance

*No products are associated with this question.*

Answer by Johann
on 8 Aug 2012

Edited by Johann
on 16 Aug 2012

Accepted answer

Hello Usman Ali,

the following code works for me:

% scennum_fav is the favored scenario number after reduction scennum_fav=250; % Initial scenario numbers, my matrix to be reduced is CO2_scen [1000x234] scennum_int=size(CO2_scen,1); % In the beginning all scenarios are equiprobable scenprob=ones(scennum_int,1)*1/scennum_int;

% scenario reduciton algorithm while size(CO2_scen,1)>scennum_fav

% Calculate euclidean distances

dist_mat = queeze(sqrt(sum(bsxfun(@minus,CO2_scen,reshape(CO2_scen',1,size(CO2_scen,2),size(CO2_scen,1))).^2,2))); dist_mat(~dist_mat)=inf; row_min = min(dist_mat); min_value = min (row_min);

[row column]= find(dist_mat==min_value); % shows the position of the minimum element, r= which row and c= which column

% Remove scenario with smallest distance and adding its probability to reference scenario (the scenario to be removed is independent of its probability)

if scenprob(row(1))>scenprob(row(2)) CO2_scen(row(2),:)=[]; scenprob(row(1),:)=scenprob(row(1),:)+scenprob(row(2),:); scenprob(row(2),:)=[]; elseif scenprob(row(1))<scenprob(row(2)) CO2_scen(row(1),:)=[]; scenprob(row(2,:))=scenprob(row(2,:))+scenprob(row(1,:)); scenprob(row(1),:)=[]; else CO2_scen(row(2),:)=[]; scenprob(row(1,:))=scenprob(row(1,:))+scenprob(row(2,:)); scenprob(row(2),:)=[]; end end

I hope this was helpful even though your question was a few month ago. Johann

## 0 Comments