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Index exceeds the number of array elements

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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 Comments
Rik
Rik on 28 Jan 2019
What is your goal with this code? It might be possible to do this in a much faster and cleaner way.
rpid
rpid on 28 Jan 2019
Edited: rpid on 28 Jan 2019
I need to construct this:
x(i) h(k)=sum(x(i):x(i+2))
1
2
3 6
4
5
6 24
7
8
9 42
10
11
12 60
13
14
15 78
16
17
18 51
19
20
21 60
22
23
24 69
25
26
27 78
28
29
30 87
I need a sequence as presented in h(k)=sum(x(i):x(i+2))...
The 2nd column is a sum of three elements of the 1st column.

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Accepted Answer

madhan ravi
madhan ravi on 28 Jan 2019
Edited: madhan ravi on 28 Jan 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

More Answers (6)

madhan ravi
madhan ravi on 28 Jan 2019
Edited: madhan ravi on 28 Jan 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
  1 Comment
rpid
rpid on 28 Jan 2019
Thank you! It It works perfectly.
Thank you all for attention.

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Rik
Rik on 28 Jan 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.
  2 Comments
rpid
rpid on 28 Jan 2019
Thank you. It works perfectly also.
rpid
rpid on 28 Jan 2019
Thank you. It works perfectly also.

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madhan ravi
madhan ravi on 28 Jan 2019
Edited: madhan ravi on 28 Jan 2019
x=1:30;
Result=squeeze(sum(reshape(x,3,1,[]))) % without transpose

rpid
rpid on 28 Jan 2019
Edited: rpid on 28 Jan 2019
Thank you again. All answers worked.
I just don't know how can I apply for a sequence not divisible by other numbers. I have a sequence of 1118 elements and I need to sum the same sequence in 3 to 3, 6 to 6, 9 to 9, 12 to 12, 18 to 18, 24 to 24, 48 to 48...
The same procedure, but applied in these:
x=1:30;
Result=squeeze(sum(reshape(x,3,1,[]))) % without transpose
I have the input data a matrix 1:1118 and 2 columns.
  3 Comments
Prasanth Warrier
Prasanth Warrier on 11 Nov 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?

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Suganthi D
Suganthi D on 15 Mar 2022
Edited: Walter Roberson on 15 Mar 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 Comments
dugasa getachew
dugasa getachew on 9 May 2022
Edited: dugasa getachew on 9 May 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
Rik on 9 May 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.

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Jaqueline Hernández
Jaqueline Hernández on 28 Nov 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 Comments
Jaqueline Hernández
Jaqueline Hernández on 28 Nov 2022
En la primera linea me marca el error que el indice supera el numero de matrices
disp ('introducir valores en S.I. ');
DGM
DGM on 28 Nov 2022
As Walter mentioned, check that you don't have a variable called disp.
% disp is a function
disp('introducir valores en S.I. ');
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. ');
Index exceeds the number of array elements. Index must not exceed 3.

'disp' appears to be both a function and a variable. If this is unintentional, use 'clear disp' to remove the variable 'disp' from the workspace.
% 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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