calculate values from different matrices
    4 Ansichten (letzte 30 Tage)
  
       Ältere Kommentare anzeigen
    
    babis
 am 28 Mai 2014
  
    
    
    
    
    Beantwortet: Roger Wohlwend
      
 am 28 Mai 2014
            I have a matrix U (n*v). I want for every column to have a matrix that has the number of the non-zero rows. For example U= [1 1 0; 2 0 2; 4 3 0; 4 0 0; 0 5 1; 3 1 3; 1 0 5; 0 2 1] will return for v1=1 [1 2 3 4 6 7], v2=2 [1 3 5 6 8], v3=3 [2 5 6 7 8]. Then, we have another matrix K (n*n) and we want for every new matrix we created before(v1 v2 v3) , to take the 5 nearest to zero values. for example [0.1 0.5 0.2 0.25 0.3 -0.1 0.4 0.35; 0.3 0.3 0.2 0.25 0.3 -0.1 0.4 0.35; 0.15 0.5 0.4 0.25 0.3 -0.1 0.4 0.35; -0.4 0.5 0.2 0.6 0.3 -0.1 0.4 0.35; 0.3 0.5 0.2 0.25 0.12 -0.1 0.4 0.35; 0.15 0.5 0.2 0.25 0.3 -0.1 0.4 0.35; 0.3 0.5 0.2 0.25 0.3 -0.1 0.213 0.35; 0.125 0.5 0.2 0.25 0.3 -0.1 0.4 0.12]
for row 1 is [1 3 4 6 7]
then, we want to create a new matrix A where every element is the sum of each absolute value of K multiplied with the respective values of U, and the sum will be divided by the sum of the 5 absolute values from K. For example: A(1,1)= {1*0.1+4*0.2+4*0.25+3*|-0.1|+1*0.4}/(0.1+0.2+0.25+|-0.1|+0.4)
0 Kommentare
Akzeptierte Antwort
  Roger Wohlwend
      
 am 28 Mai 2014
        Task 1:
m = size(U,2);
v = cell(m,1);
for k = 1 : m
   v{k} = find(U(:,k) > 0);
end
Task 2: I don't understand what you want to do. v1, v2 and v3 are vectors of different lengths. How do you use them on the matrix K?
0 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
				Mehr zu Linear Algebra finden Sie in Help Center und File Exchange
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!

