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

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mhowell
mhowell 2013-5-4
回答： deeba naqvi 2020-10-15
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

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2013-5-4
编辑：Azzi Abdelmalek 2013-5-4
After one itteration
for i=1:4
x = x(i)
the length of x becomes equal to 1, then x(2) does not exist

更多回答（3 个）

Iman Ansari
Iman Ansari 2013-5-4
Hi.
You change your x in first loop :
x = x(i) ====> x=22.5
after this x became a number.
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Valentina Marincevic Petracic
Valentina Marincevic Petracic 2017-1-5
编辑：Stephen23 2017-1-5
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

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Aaina
Aaina 2018-8-7
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
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Aaina
Aaina 2018-8-7
编辑：Aaina 2018-8-7
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

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deeba naqvi
deeba naqvi 2020-10-15
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);

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