how can i name each ans in a for loop?

Nb=6;

L =[1 2 0.1 0.2 0.02

1 4 0.05 0.2 0.02

1 5 0.08 0.3 0.03

2 3 0.05 0.25 0.03

2 4 0.05 0.1 0.01

2 5 0.1 0.3 0.02

2 6 0.07 0.2 0.025

3 5 0.12 0.26 0.025

3 6 0.02 0.1 0.01

4 5 0.2 0.4 0.04

5 6 0.1 0.3 0.03]

%-------------------------------Program strat her--------------------------

nl = L(:,1); nr = L(:,2); R = L(:,3);

X = L(:,4); Bc = j*L(:,5);

nbr=length(L(:,1)); nbus = max(max(nl), max(nr));

for i=1:Nb

rowsToSum = L(:,1) == i ;

theSum = sum(L(rowsToSum, 4));

disp(theSum)

end

for example

0.7000=a

0.8500=b

0.3600=c

0.4000=d

0.3000=e

0=f

[Stephen23]

"how can i name each ans in a for loop?"

While that is certainly possible, doing so will force you into writing slow, inefficient, complex, obfuscated, buggy code that is difficult to debug. In contrast using indexing is simple, efficient, and easy to use.

[arian hoseini]

i get what u saying so if that's not an option....so help me with this

i need to create this matrix...

[x11 x12 ...]......x11 is rowsToSum = L(:,1) == 1 ;

and x12 is rowsToSum = L(:,1) == 1 ;rowsToSum = L(:,2) == 2 ;

and this continues...

(this is the reason im asking for naming ans in a loop)

[Steven Lord]

Make X a matrix or a cell array depending on whether you want to store how many elements match 1, 2, etc. or you want to store the elements themselves. One easy way to do the former, if all of the bins you want are integer values, is histcounts.

% Sample data
x = randi(10, 1, 100);
% Bin the data
[d, edges] = histcounts(x, 'BinMethod', 'integer');
% Display the results
results = table(d.', edges(1:10).', edges(2:11).', ...
    'VariableNames', {'counts', 'left edge', 'right edge'})

results = 10×3 table
    counts    left edge    right edge
    ______    _________    __________

      12         0.5           1.5   
      10         1.5           2.5   
      10         2.5           3.5   
       8         3.5           4.5   
      11         4.5           5.5   
      11         5.5           6.5   
      14         6.5           7.5   
       8         7.5           8.5   
       6         8.5           9.5   
      10         9.5          10.5   

Let's double-check with the simpler approach.

howManyFours = sum(x == 4) % Compare with the table results, between edges 3.5 and 4.5

howManyFours = 8

Instead of referring to a variable named d4 access the 4th element of d, d(4).

howManyFoursApproach2 = d(4)

howManyFoursApproach2 = 8

[arian hoseini]

thanks.

Answer 1

[Voss]

You are checking which rows of L have Nb (i.e., 6) in the first column:

rowsToSum = L(:,1) == Nb ;

Instead you should check which rows of L have i in the first column:

rowsToSum = L(:,1) == i ;

And to store them all do this:

theSum = zeros(1,Nb);
for i=1:Nb
    rowsToSum = L(:,1) == i ;
    theSum(i) = sum(L(rowsToSum, 4));
end
disp(theSum)

[arian hoseini]

yeah i noticed that sorry for wasting your time i have another question how can i name them?

[Steven Lord]

Can you define variables with numbered names like X1, X2, X3, ... ? Yes.

Should you do this? Generally we recommend against it. See that page for alternatives you should use instead.

[arian hoseini]

not helping i read that page before but still dont know what to do....if u can tell me exactly what to do i'll appreciate that..

[Voss]

Based on your comment here, it sounds like you want to perform each sum of rows of L based on their values in column 1 and column 2, which you can do like this.

Nb=6;
L =[1 2 0.1 0.2 0.02
    1 4 0.05 0.2 0.02
    1 5 0.08 0.3 0.03
    2 3 0.05 0.25 0.03
    2 4 0.05 0.1 0.01
    2 5 0.1 0.3 0.02
    2 6 0.07 0.2 0.025
    3 5 0.12 0.26 0.025
    3 6 0.02 0.1 0.01
    4 5 0.2 0.4 0.04
    5 6 0.1 0.3 0.03];

x = zeros(Nb);
for i=1:Nb
    for j=1:Nb
        rowsToSum = L(:,1) == i & L(:,2) == j;
        x(i,j) = sum(L(rowsToSum, 4));
    end
end
disp(x)

         0    0.2000         0    0.2000    0.3000         0
         0         0    0.2500    0.1000    0.3000    0.2000
         0         0         0         0    0.2600    0.1000
         0         0         0         0    0.4000         0
         0         0         0         0         0    0.3000
         0         0         0         0         0         0

So the element of x in the ith row, jth column is the sum of the 4th column of L over rows that start with [i j].

[arian hoseini]

thank u.your answer was really helpfull but one problem x(1,1) is the sum of all X's whose nl=1 and x(2,2),....

[arian hoseini]

and x(2,1) is 0.2 ....