Index exceeds the number of array elements
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rpid
el 28 de En. de 2019
Comentada: DGM
el 28 de Nov. de 2022
Dear all. How can I fix this: "Index exceeds the number of array elements"?
data=[1:30]';
x=data(:,1);
n=length(data);
for i=1:n-2
for k=1:n-2
h(k)=sum(data((i):(i+2)));
k=k+1;
i=i+3;
end
end
Thank you all!
3 comentarios
Rik
el 28 de En. de 2019
What is your goal with this code? It might be possible to do this in a much faster and cleaner way.
Respuesta aceptada
madhan ravi
el 28 de En. de 2019
Editada: madhan ravi
el 28 de En. de 2019
One way:
x=(1:30).';
[m,n]=size(x);
N=3;
Result=zeros(m/N,n); % preallocate
for k = 1:m/N
Result(k,n)=sum(x(k*N-(N-1):k*N));
end
Result
Gives:
Result =
6
15
24
33
42
51
60
69
78
87
0 comentarios
Más respuestas (6)
madhan ravi
el 28 de En. de 2019
Editada: madhan ravi
el 28 de En. de 2019
Another way:
x=(1:30).';
[m,n]=size(x);
N=3;
C=mat2cell(x,repmat(N,1,m/N));
Result=cellfun(@sum,C)
Gives:
Result =
6
15
24
33
42
51
60
69
78
87
Rik
el 28 de En. de 2019
This would also work:
x=(1:30)';
Result=sum(reshape(x,3,[]))';
The transposes are not needed, but are only there to keep input and output shape consistent with the other methods by Madhan.
madhan ravi
el 28 de En. de 2019
Editada: madhan ravi
el 28 de En. de 2019
x=1:30;
Result=squeeze(sum(reshape(x,3,1,[]))) % without transpose
0 comentarios
rpid
el 28 de En. de 2019
Editada: rpid
el 28 de En. de 2019
3 comentarios
Prasanth Warrier
el 11 de Nov. de 2020
I have an array created with this command
t = [0:0.01:0.99 1:0.1:9.9 10:1:100];
this creates 281 columns
I want to know difference between each columns
like time interval between 2nd and 1st colulmn and between 2nd and 3rd like wise.
How do i go about it?
Suganthi D
el 15 de Mzo. de 2022
Editada: Walter Roberson
el 15 de Mzo. de 2022
good Morning professors,
can anyone help me to sort out this problem..
Index exceeds the number of array elements (127).
Error in fitness_process2_12 (line 83)
f1tmp(km)=elec_price(km)*(pdis_charget(km)/1e3)*delt; % objective function 1
Error in
minlp_process_12>@(x)fitness_process2_12(x,EVSE1_arr_time,EVSE1_leave_time,EVSE2_arr_time,EVSE2_leave_time,EVSE3_arr_time,EVSE3_leave_time,finaltime,socintij1,socintij2,socintij3,data_pass)
(line 14)
func=@(x) fitness_process2_12(x,EVSE1_arr_time,EVSE1_leave_time,...
Error in fmincon (line 567)
initVals.f = feval(funfcn{3},X,varargin{:});
Error in minlp_process_12 (line 29)
final_best_value=fmincon(func,(intial_solu),adatamat,bdatamat,[],[],lower_lmt,upper_lmt,[],options); %%Call fmincon ----
with the fval output to obtain the value of the objective function at the solution.
Error in MARCH (line 61)
[finalres]=minlp_process_12(min_val1,max_val1,EVSE1_arr_time,EVSE1_leave_time,...
Caused by:
Failure in initial objective function evaluation. FMINCON cannot continue.
5 comentarios
dugasa getachew
el 9 de Mayo de 2022
Editada: dugasa getachew
el 9 de Mayo de 2022
% Measure obje performance of the (GA based parameter optimization)
function IEA= EvalObj(x)
th=x(1); dis=x(2);interDis=x(3);fr=x(4);
radius = 0.31;
interRadius = 0.015;
omega = 2*pi*fr;
permVac = 4*pi* 10^(-7) ;
permRel = 1500;
perm = permVac*permRel;
cond = 10.1^6;
Current = 10; % Coil Currnet
alpha = 10000;
Lamb =235;
h = 11;
T = 200 ;
Ta=25;
domain = [0 0.29 0 th];
r1= interRadius: interDis:radius;
alpha1 = @(x,r,z) sqrt(x.^2 + 1j*omega*perm*cond) ;
num = @(x,r,z) (exp(-alpha1(x,r,z).*(th-z)) .*(alpha1(x,r,z)-x) + exp( alpha1(x,r,z).*(th-z)) .*(alpha1(x,r,z) + x));
den = @(x,r,z) exp(alpha1(x,r,z).*th).*(alpha1(x,r,z) + x).^2 - exp(-alpha1(x,r,z).*th) .*(alpha1(x,r,z)-x).^2;
H = 0;
IEA=0;
for i = 1:28
tic
i
f = @(x,r,z) besselj(1,x*r1(i)).*besselj(1,x.*r).*exp(-x*dis).*x;
Temp = @(r,z) sum(chebfun(@(x) f(x,r,z) .*num(x,r,z) ./ den(x,r,z), [ 0 alpha]));
Flux = chebfun2(@(r,z) perm*Current*r1(i)*Temp(r,z), domain,'vectorize on');
H = H + Flux;
toc
J= 1j*omega*cond*H;
Ind =abs(J);
x = chebfun2(@(x,y) x, domain)
N = chebop2(@(x,y,u) Lamb.*x.*diffx(u,2) + Lamb.*diffx(u,1) + Lamb.*x.*diffy(u,2) , domain);
N.lbc =@(x,u) diff(u);
N.rbc = @(x,u) diff(u,1) +h*(u - Ta) ./Lamb; % z- direction
N.ubc= @(x,u) diff(u,1) +h*(u - Ta) ./Lamb; % r - direction
N.dbc = @(x,u) diff(u);
s = x.*Ind;
u=N \-s;
%u=u(:,domain(4));
IEA=IEA+(u-T)/T;
end
%%%%%%%%%%%%%%%%
Index exceeds the number of array elements (1).
Error in EvalObj>@(r,z)perm*Current*r1(i)*Temp(r,z) (line 28)
Flux = chebfun2(@(r,z) perm*Current*r1(i)*Temp(r,z), domain,'vectorize
on');
what is problem i tried many times but couldn't get the solution
Rik
el 9 de Mayo de 2022
@dugasa getachew: have a read here and here. Then post your question to a separate thread. It will greatly improve your chances of getting an answer.
Jaqueline Hernández
el 28 de Nov. de 2022
¿Cómo puedo solucionar esto: "El índice supera el número de elementos de la matriz"?
disp ('introducir valores en S.I. ');
b= input('¿cual es el ancho del canal? ');
yl=input('¿cual es el valor de yl? ');
disp ('dz positivo si es ascendente ');
dz=input ('¿cual es el valor de dz? ');
Q= input('¿cual es el valor del gasto? ');
Al= b*yl;
vl_2=(Q/Al)^2;
cvl=vl_2/(2*9.81);
El=yl+cvl;
Ea= El-dz
a=(Q^2/(2*9.81*b^2));
r=roots([1,-Ea,0,a])
y=(.0000001:.01:(2*y1));
E= y+(Q^2./((2*9.81)*(b*y).^2));
plot (E,y,Ea,r,'rd',El,y1,'go')
axis([0 5 0 5])
title ('GRAFICA E vs y')
xlabel ('Energia, E (m)')
Ylabel ('Tirante, y (m)')
grid
3 comentarios
Jaqueline Hernández
el 28 de Nov. de 2022
En la primera linea me marca el error que el indice supera el numero de matrices
disp ('introducir valores en S.I. ');
DGM
el 28 de Nov. de 2022
As Walter mentioned, check that you don't have a variable called disp.
% disp is a function
disp('introducir valores en S.I. ');
% disp is now a variable (e.g. a numeric array)
disp = [1 2 3];
% now you're trying to index into an array using a character vector
% note that the linter will now highlight this line since it's an
% addressing expression with no output
disp('introducir valores en S.I. ');
% which is implicitly handled as numeric indices
disp([105 110 116 114 111 100 117 99 105 114 32 118 97 108 111 114 101 115 32 101 110 32 83 46 73 46 32]);
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