Why doesn't the loop in my for loop work?

Hi!

Im trying to get a for-loop to work with an if-statement, but I can't seem to get it to loop at all. Please help!

The snippet of the code that doesn't loop is bellow and Im sorry that the names are in swedish but it shouldn't be a problem...

The "besparingslistan" is a list with 3 coulmns and I want the for loop "f" to loop through all rows in it but only the first column, the same with "a" but only the second column. I don't know if I'm doing the for loops wrong in the beginning or if it just isn't working with "f=f+1" and "a=a+1" at the end.

If the "if-statement" is false, I want the loop to continue to the next row in the "besparingslistan" and therefore don't save anything in the "turkonfig".

for f = 1:Besparingslistan(:,1)
    for a = 1:Besparingslistan(:,2)
        if Kund_i_tur(f)~= Kund_i_tur(a)
            Turkonfig = {f a};
        else 
            f=f+1;
            a=a+1;
        end
    end
end

Answer 1

[Sai Sri Harsha Vallabhuni]

Hey,

Below is modified code which achieves what you are tying to do

for f = Besparingslistan(:,1)
    for a = Besparingslistan(:,2)
        if Kund_i_tur(f)~= Kund_i_tur(a)
            Turkonfig = {f a};
        end
    end
end

Hope this helps.

[Amanda S]

Hi! Thank you but it still doesnt work, or at least not the way that i thought it would. Perhaps there's a mistake where Turkonfig saves the values in a cell array? Isn't it just overwriting the old loop value when it just says Turkonfig = {f a}? Is there some way to save the f and a in seperate rows after each loop?

Answer 2

[Steven Lord]

for f = 1:Besparingslistan(:,1)

This doesn't do what you think it does. Consider:

A = [20 1 2; 2 3 4; 3 4 5; 4 5 6];
for f = 1:A(:, 1)
    disp("A(" + f + ", 1) is " + A(f, 1))
end

You will receive an error when f is 5, since A only has 4 rows. Instead of using the values in A to set the limits of your for loop, use the size of A.

A = [20 1 2; 2 3 4; 3 4 5; 4 5 6];
numberOfRows = size(A, 1);
for f = 1:numberOfRows
    disp("A(" + f + ", 1) is " + A(f, 1))
end

Also, making changes to the loop variable inside the loop is discouraged. It doesn't do what you want anyway; any changes you made to that loop variable are thrown away when the next iteration of the loop starts and MATLAB takes the next value in the vector.

for k = 1:10
    disp("k before changing is " + k)
    k = 42;
    disp("k after changing is " + k)
end

[Amanda S]

Hmm alright, yeah I suspected that I was doing something wrong like that. But I still don't quite understand how I would do it instead? Lets say that I use the size of "Besparingslistan" instead, but then I get an error saying "Index exceeds the number of array elements (13)." What does that mean?

The "Besparingslistan" is a list with the size 169x3 built up with different combinations of 13 different customers in the first two columns and a savingsvalue in the third column. While Kund_i_tur and Turkonfig are column vectors with the customers, aka (1:1:13). Could it be something like the example with "You will receive an error when f is 5, since A only has 4 rows." ?

f = size(Besparingslistan,1);
a = size(Besparingslistan,2);

for forstatur = 1:f
    for andratur = 1:a
        if Kund_i_tur(f)~= Kund_i_tur(a)
           Turkonfig = {f a};
        end
    end
end