Attempted to access x(2); index out of bounds because numel(x)=1. But I'm using a matrix?

I'm just trying to execute a simple for loop, but I keep getting this error "Attempted to access x(2); index out of bounds because numel(x)=1.

Error in Quiz (line 11) x = x(i)" I don't understand what the error means. Any help would be appreciated!

x = [22.5 45 67.5 90]

 for i=1:4
    x = x(i)
    a = a(x)
    c = c(x)
    alpha = alpha(x)
    mu(a,c)
end

Answer 1

[Azzi Abdelmalek]

After one itteration

for i=1:4
  x = x(i)

the length of x becomes equal to 1, then x(2) does not exist

Answer 2

[Iman Ansari]

Hi.

You change your x in first loop :

x = x(i) ====> x=22.5

after this x became a number.

[mhowell]

That makes sense. So how do I go about calling all the values of the matrix into my for loop?

[Iman Ansari]

Use another variable:

y = [22.5 45 67.5 90]
y(2)
n = [1 3 5 7]
for i=1:4
  x = y(i)
  a = a(x)
  c = c(x)
  alpha = alpha(x)
  mu(a,c)
end

[mhowell]

Thanks!!!

[Pradeep Kumar R]

what should i do if i get the same error while using a if loop inside a for to plot a wave

[Valentina Marincevic Petracic]

same question:

function [QRS] = AF2(EKG)

QRS=zeros(1,8191);

a=max(EKG);
Xth=0.4*a; %amplitude threshold
Y0=zeros(1,8191);
Y1=zeros(1,8191);
Y2=zeros(1,8191);

for n=1:8191
    if (EKG(n)>0)
        Y0=EKG(n);
    elseif (EKG(n)<0)
        Y0=-EKG(n);
    end
end

%NP filtar:
for k=1:8191
    y=Y0(k);
    if ( y>= Xth)
        Y1(k)=y;
    else
        Y1(k)=Xth;
    end 
end

%prva derivacija:
for n=2:8190
    Y2= Y1(n+1)- Y1(n-1);
end

%detekcija qrs-a:
for i=0: 8191
    if((Y2(i))>0.7)
        QRS(i)=1;
    end
end

end

Answer 3

[Aaina]

How can I solve this problem?

Attempted to access C(38,1); index out of bounds because size(C)=[37,38].

Error in Without_DG (line 94) if ((C(f,i)==-1)&&(k==1));

%% MATLAB Program

       k=1;
       for i=1:no
           if ((C(f,i)==-1)&&(k==1));
                f=i;
                g(j,e)=i;
                e=e+1;
                k=3;
           end            
       end

[Stephen23]

"How can I solve this problem?"

Problem: trying to access array elements that don't exist throws an error.

Solution: don't try to access array elements that don't exist.

Maybe you need to check your input data better. Maybe your algorithm logic is incorrect. Maybe your wrote your code with a bug. We don't know because one code excerpt is not enough for us to know what you are doing and why. So find out what you did wrong and fix it.

[Aaina]

Alright.... Thank you so much.... the input data is correct, but the code was incorrect, i tried to change the data by using the same code which need to access different network of distribution system... which from radial system into mesh system which the network more complex

[Aaina]

I want the code can access mesh system by just adding the mesh data, but it seems doesnt work

Answer 4

[deeba naqvi]

Even I am getting the same error..plz some body help.....The programm code is.....

lb = [0,0,0,-inf];

Aeq = [1 1 1 0];

beq = 1;

x0 = [0.25 0.25 0.25 8];

fun = @(x) -x(4);

A = [];

b = [];

ub = [];

nonlinear = @Alteredplay2;

nonlcon = nonlinear(x);

[X, FVAL] = fmincon(fun, x0, A, b, Aeq, beq, lb, ub, nonlcon)

function [C, Ceq] = Alteredplay2(x)

%x0 = [0.25 0.25 0.25 2];

C(1)= x(4)-((1-(5.8/8))^(0.5)*(0.1/8)^(0.5)*(1-(4.4/8))^(0.5)*(1.7/8)^(0.5))^x(1)*((1-(2/8))^(0.5)*(2.9/8)^(0.5)*(1-(1.4/8))^(0.5)*(5.8/8)^(0.5))^x(2)*((1-(3.4/8))^(0.5)*(0.1/8)^(0.5)*(1-(5.2/8))^(0.5)*(2.7/8)^(0.5))^x(3);

C(2)= x(4)-((1-(6.1/8))^(0.5)*(0.6/8)^(0.5)*(1-(3.8/8))^(0.5)*(1.7/8)^(0.5))^x(1)*((1-(2.4/8))^(0.5)*(3.1/8)^(0.5)*(1-(1.3/8))^(0.5)*(5.6/8)^(0.5))^x(2)*((1-(7.1/8))^(0.5)*(0.3/8)^(0.5)*(1-(3.6/8))^(0.5)*(0.8/8)^(0.5))^x(3);

Ceq= [];

end

and the error is....

Attempted to access x(4); index out of bounds because

numel(x)=2.

Error in Alteredplay2 (line 3)

C(1)=

x(4)-((1-(5.8/8))^(0.5)*(0.1/8)^(0.5)*(1-(4.4/8))^(0.5)*(1.7/8)^(0.5))^x(1)*((1-(2/8))^(0.5)*(2.9/8)^(0.5)*(1-(1.4/8))^(0.5)*(5.8/8)^(0.5))^x(2)*((1-(3.4/8))^(0.5)*(0.1/8)^(0.5)*(1-(5.2/8))^(0.

Error in ExceptPlay2Altered (line 10)

nonlcon = nonlinear(x);