how to call a specific variable in matrix inside loop process..??

I have a little problem with my matrix. Perhaps you interested with this..

I have matrix like below :

   xdata = [ 1     2    0.0192   
             1     3    0.0452   
             2     4    0.0570   
             3     4    0.0132   
             2     5    0.0472   
             2     6    0.0581   
             4     6    0.0119   
             5     7    0.0460   
             6     7    0.0267   
             6     8    0.0120   
             6     9    0.0      
             6     10   0.0      
             9     11   0.0   ];

I want to call a variable in that matrix one by one following my loop order. For example :

Q = 0;
K = 0;
n = length(xdata);
i = xdata(:,1);
j = xdata(:,2);
for bb = 1:n,
   for cc = 1:n,
       c1 = xdata(:,1);
       c2 = xdata(:,2);
       if c1 == bb,
           if c2 == cc,
           K = linedata(:,3);
           end
       end
       Q = Q + (K*100);
   end
end

so when bb = 1 and cc = 2, K will call xdata(:,3) in line 1 = 0.0192

when bb = 1 and cc = 3, K will call xdata(:,3) in line 2 = 0.0452

when bb = 6 and cc = 7, K will call xdata(:,3) in line 9 = 0.0267, and so on...

My code above are wrong and I feel really dizzy when coding it, maybe you can help me with this..

anyone has the solution..?

thanks...

[Cedric]

It is difficult to understand what you tried to do looking at a code that doesn't work. Could you explain a bit more about what you want to achieve with this array (and not how you tried to process it)?

[Noru]

ok, i want to solve a simple equation of Q, Q will loop with bb = 1:n, and inside Q equation i want to call K variables like i've explain before..

i hope this explanation are easy to understand..

[Cedric]

It is not actually. What is Q and what does it mean to solve an equation of Q? Just to explain why its is difficult to understand, in your code..

• There is this linedata that is defined nowhere.
• Assuming that it is some array of at least 3 columns, when you write K = linedata(:,3);, K is a column vector with the same number of rows as linedata.
• Same with c1 and c2; these are column vectors with the same number of rows as xdata.
• Your IF statements are not valid a priori, because e.g. c1==bb will generate an array of logicals of the same size as c1, and using this in an IF statement will certainly not produce the effect that you were trying to implement.
• Etc.

Could you write exactly what you are trying to achieve for a few terms without using the loop?

[Noru]

ok Cedric, my data above is a linedata from one of a "electrical power bus data system", and now I am studying it with optimization process. To get the best result of the optimization i have to evaluate it. I have a complicated equation and Q are a simple equation that i've been made to represent my evaluation equation.

In my real equation, i have a problem when calling 'xdata(:,3)', because 'xdata(:,3)' have to called when bb = 1:n and cc = 1:n like an example below :

for bb = 1:n,
   for cc = 1:n,
      Q = Q+sum(K*(A(bb)-B(cc)));
   end
end

for an additional data :

A =

    0.9148
    0.4772
    0.9162
    0.4969
    0.9276
    0.7037
    0.6946
    0.0046
    0.5402
    0.4130

B =

    0.6281
    0.5435
    0.2160
    0.5521
    0.8627
    0.0049
    0.1251
    0.7751
    0.0204
    0.1426

so when bb = 1 and cc = 2, Q = sum(0.0192*(0.9148-0.5435)) and so on...

but when bb = 1 and cc = 1, Q = 0 because there is no K data called in 'xdata'.

is this clear enough..? I am sorry to make you confuse..

[Cedric]

Ok, it is starting to get more clear. In your initial question, the line which defines K defines it as a column vector. In your last comment, it as actually a scalar (the element of xdata(:3) on the row which has its first two elements matching bb and cc). In your last comment, you also write

 Q = Q+sum(K*(A(bb)-B(cc)));

in a nested FOR loop. If everything in this equation is scalar, why using SUM? Isn't it instead the following?

 Q = Q + K * (A(bb) - B(cc)) ;

with K a number/scalar (0.0192) and A(bb) and B(cc) scalars as well?

Finally, xdata(2,end) is 11, when B has only 10 elements; how should this be managed?

Answer 1

[Cedric]

Here is an answer based on what I understood, and a truncated version of xdata in which I removed the last row that contains an invalid index (11) for indexing B.

It seems to me that what you want to do can be reduced to one vector operation as follows:

 >> Q = sum( xdata(:,3) .* (A(xdata(:,1)) - B(xdata(:,2))) )
 Q =
    0.1059

where the SUM operates along dimension 1 (by default). I will discuss it a little more afterwards, but let's check against a corrected version of your double, nested FOR loop:

 >> n = size(xdata, 1) ; 
 >> Q = 0 ;
 >> for bb = 1 : n
       for cc = 1 : n
          id = xdata(:,1)==bb & xdata(:,2)==cc ;
          if any(id)
             Q = Q + xdata(id,3) * (A(bb) - B(cc)) ;
          end
       end
    end

 >> Q
 Q =
    0.1059

which seems to match to concise version. In these FOR loops, finally only a few iterations lead to a match of indices. If you look at what we are doing in the concise version,

 >> A(xdata(:,1))
 ans =
    0.9148
    0.9148
    0.4772
    0.9162
    0.4772
    0.4772
    0.4969
    0.9276
    0.7037
    0.7037
    0.7037
    0.7037

is a vector of elements of A that correspond to these matches. Same for B(xdata(:,2)). By subtracting them, we get in one shot all these factors A(bb)-B(cc), that we can multiply element-wise by all the K's ( xdata(:,3) ). Finally, we take the sum of these products.

Test this/these solution(s) little bit by little bit to get full understanding. Let me know if I didn't understand well what you need, or if there is anything that you don't understand.

[Noru]

ok thank you Cedric..

your solution really solve my problem..

thanks again..

Answer 2

[Archit]

a way (maybe a better way exists, but i am not very proficient in matlab yet) to do that is

a=find(xdata(:,1)==2);
% a=[3 5 6];
b=find(xdata(a,2)==4);
% b=[1]
A=xdata(a,1);
B=A(b,3);
% B=0.0570;

basically u filter results by columns till u get ur answer

sorry i cannot think of a more general way to do this at the moment

[Noru]

thanks for your solution, but there is an error with your coding

"Attempted to access A(1,3); index out of bounds because size(A)=[3,1]."

and i hope the solution is in 'for loop' operation..

thanks again..