Problem using if statements in for loop

I use these codes to generate a matrix inside the for loop:

kd=zeros(1,7);

kd=[0.11:0.01:0.17];

for ii=1:length(kd)

for j=1:7

kp(j)=j;

ki(j)=j;

AA(ii,j)=ki(j);

end

end

Result:

AA =

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

Then I put if statements inside the for loop to change the elements of the matrix AA. Interestingly, the result is as follows:

kd=zeros(1,7);

kd=[0.11:0.01:0.17];

for ii=1:length(kd)

for j=1:7

kp(j)=j;

ki(j)=j;

if kd(ii) == 0.11

ki(j)=100;

elseif kd(ii) == 0.12

ki(j)=200;

elseif kd(ii) == 0.13

ki(j)=300;

elseif kd(ii) == 0.14

ki(j)=400;

elseif kd(ii) == 0.15

ki(j)=500;

elseif kd(ii) == 0.16

ki(j)=600;

elseif kd(ii)==0.17

ki(j)=700;

end

AA(ii,j)=ki(j);

end

end

Result:

AA =

100 100 100 100 100 100 100

200 200 200 200 200 200 200

300 300 300 300 300 300 300

400 400 400 400 400 400 400

1 2 3 4 5 6 7

600 600 600 600 600 600 600

700 700 700 700 700 700 700

How can I solve this problem?

[KSSV]

kd=zeros(1,7);
kd=[0.11:0.01:0.17];
for ii=1:length(kd)
    for j=1:7
        kp(j)=j;
        ki(j)=j;
        
        AA(ii,j)=ki(j);
    end
end 

The above can be just achieved with:

kd = repmat(1:7,7,1)

kd = 7×7

1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7

[Serdar Ethem Hamamci]

thanks KSSV, it was very useful for me.

Answer 1

[Jan]

Welcome to the world of numerical maths. See: FAQ: Why is 0.3-0.2-0.1 not equal to zero?

The standard example is:

any((0:0.1:1) == 0.3)    % FALSE!!!

ans = logical
   0

Most floating point numbers cannot be represented exactly in the binary format, they are stored in in the IEEE754 standard. This is the format used in Matlab and on all modern CPUs.

This means, that the IF command is working as expected, but kd(ii) == 0.15 is not suitable:

format long g
kd = 0.11:0.01:0.17;
kd.' - 0.15

ans = 7×1

-0.04 -0.03 -0.02 -0.00999999999999998 2.77555756156289e-17 0.01 0.02

Do you see it?

By the way, [ ] is the concatenation operator in Matlab. In [0.11:0.01:0.17] you concatenate the vector 0.11:0.01:0.17 with nothing, so this is just a waste of time.

[Serdar Ethem Hamamci]

Thanks jan, it was very useful for me. I will use the following codes for now. If anyone has a different short code suggestion, I can take it.

kd=zeros(1,7);

kd=[0.11:0.01:0.17];

for ii=1:length(kd)

for j=1:7

kp(j)=j;

ki(j)=j;

if abs(kd(ii)-0.11)<0.001

ki(j)=100;

elseif abs(kd(ii)-0.12)<0.001

ki(j)=200;

elseif abs(kd(ii)-0.13)<0.001

ki(j)=300;

elseif abs(kd(ii)-0.14)<0.001

ki(j)=400;

elseif abs(kd(ii)-0.15)<0.001

ki(j)=500;

elseif abs(kd(ii)-0.16)<0.001

ki(j)=600;

elseif abs(kd(ii)-0.17)<0.001

ki(j)=700;

end

AA(ii,j)=ki(j);

end

end

[Jan]

Some simplifications:

% kd = zeros(1,7);  % Omit this, because it is a waster of time
% You define kd but overwrite it in the following line.

% kd = [0.11:0.01:0.17];
kd = 0.11:0.01:0.17;  % As explained already:
% a:b:c is a vector already. There is no need to concatenate it with
% nothing.

AA = zeros(length(kd), 7);   % Pre-allocation!!!
for ii = 1:length(kd)
    % for j=1:7  % No need to do this repeatedly
        % kp(j)=j;  Omit this 
        % ki(j)=j;  Omit this 
        
        if abs(kd(ii)-0.11)<0.001
            ki=100;   % No need to use (j) as index
        elseif abs(kd(ii)-0.12)<0.001
            ki=200;
        elseif abs(kd(ii)-0.13)<0.001
            ki=300;
        elseif abs(kd(ii)-0.14)<0.001
            ki=400;
        elseif abs(kd(ii)-0.15)<0.001
            ki=500;
        elseif abs(kd(ii)-0.16)<0.001
            ki=600;
        elseif abs(kd(ii)-0.17)<0.001
            ki=700;
        end
        AA(ii, :) = ki;   % Set the complete row to ki
    % end
end