Attempted to access x(2); index out of bounds because numel(x)=1. But I'm using a matrix?
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mhowell
el 4 de Mayo de 2013
Respondida: deeba naqvi
el 15 de Oct. de 2020
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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Respuesta aceptada
Azzi Abdelmalek
el 4 de Mayo de 2013
Editada: Azzi Abdelmalek
el 4 de Mayo de 2013
After one itteration
for i=1:4
x = x(i)
the length of x becomes equal to 1, then x(2) does not exist
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Más respuestas (3)
Iman Ansari
el 4 de Mayo de 2013
Hi.
You change your x in first loop :
x = x(i) ====> x=22.5
after this x became a number.
5 comentarios
Pradeep Kumar R
el 25 de Feb. de 2016
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
el 5 de En. de 2017
Editada: Stephen23
el 5 de En. de 2017
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
Aaina
el 7 de Ag. de 2018
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
3 comentarios
Aaina
el 7 de Ag. de 2018
Editada: Aaina
el 7 de Ag. de 2018
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
el 7 de Ag. de 2018
I want the code can access mesh system by just adding the mesh data, but it seems doesnt work
deeba naqvi
el 15 de Oct. de 2020
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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