Index exceeds the number of array elements .

2019 3 11
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更新时间：2023 12 14
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function v = model(t,y,Z)
r=10^(-1);
d=5*10^(-1);
delta = 10^(-1);
c = 10^(2);
K = 10^(8);
C = 10^(8);
x = 1;
sigma = 10^(-2);
delta_m = 2.5*10^(-2);
ylag1 = Z(:,1);
dy = zeros(4,1);
for i = 1:6
dy(4*(i-1)+1) = r*y(4*(i-1)+1)*(1-((y(4*(i-1)+1)+y(4*(i-1)+2))/K))-d*y(4*(i-1)+3)*y(4*(i-1)+1);
dy(4*(i-1)+2) = r*y(4*(i-1)+1)*((y(4*(i-1)+1)+y(4*(i-1)+2)/K))-delta*y(4*(i-1)+3)*y(4*(i-1)+2)-delta_m*y(4*(i-1)+2);
dy(4*(i-1)+3) = c*(((ylag1(4*(i-1)+1)+ylag1(4*(i-1)+2))/C)^x)*(1-y(4*(i-1)+3));
dy(4*(i-1)+4) = 0;
end
got error in dy(4*(i-1)+1) = r*y(4*(i-1)+1)*(1-((y(4*(i-1)+1)+y(4*(i-1)+2))/K))-d*y(4*(i-1)+3)*y(4*(i-1)+1); line that Index exceeds the number of array elements .
Can someone please help me?

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Busuyi Adebayo
Busuyi Adebayo 2022-2-22
Hi Gupta. Curious what process led to this type of model formulation in delay diffenetial equations. Can you share the process?
Thanks!

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Guillaume
Guillaume 2019-3-11
编辑：Guillaume 2019-3-11

2 个投票

In the code you wrote,
for i = 1:6
So, maximum value of i is 6. It is used in your expression:
dy(4*(i-1)+1) = xxx*y(4*(i-1)+3)*xxx
So, maximum index for y is 4*(6-1)+3 == 23. So if y does not have at least 23 elements, your code will error with index exceeds the number of array elements error. We can safely conclude that it is indeed the case. The fix is to either pass an y with enough elements, or change your algorithm so that it needs less elements.
In the same vein, you declare dy as an array of 4 elements, but will assign 24 elements to it (if your i loop goes up to 6). This will not cause an error (matlab will automatically grow the array), but clearly indicates that something has not been thought out properly.
edit: As you wrote as a comment to Jan's answer, your y has 4 elements. As said, your code requires a y with 23 elements. So change your code or your y.

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parag gupta
parag gupta 2019-3-11
编辑：parag gupta 2019-3-11
for i =1
v = zeros(4,1);
v(1) = r*y(1)*(1-((y(1)+y(2))/K))-d*y(3)*y(1);
v(2) = r*y(1)*((y(1)+y(2)/K))-delta*y(3)*y(2)-delta_m*y(2);
v(3) = c*(((ylag1(1)+ylag1(2))/C)^x)*(1-y(3));
Actually these are 3 differential equations for one variant i.e v1,m1,a1 where dv1/dt = v(1) ,dm1/dt = v(2) ,da1/dt = v(2) and I solved them by using dde23 solver.
But now I want to find a solution of delay differential equations for i = 1: 6
vi(1) = r*yi(1)*(1-((y(1)+yi(2))/K))-d*yi(3)*y(1);.........................1
vi(2) = r*yi(1)*((y(1)+yi(2)/K))-delta*yi(3)*y(2)-delta_m*yi(2);...............2
vi(3) = c*(((yilag1(1)+ylag1(2))/C)^x)*(1-yi(3));.......................................3
where i is changing from 1 to 6 with different intial condition for different i.
so i wrote above code in which i got error.Could you please tell me how to edit my code so that i can get solution to system of differential equation which i have highlighted above..
thanks
Guillaume
Guillaume 2019-3-11
I don't understand what you're asking. Your code is not commented, so I've no idea what it does.
You say, I want to find a solution of delay differential equations for i = 1: 6. What does that mean? What is i, how does it relate to y?
parag gupta
parag gupta 2019-3-11
vi(1) = r*yi(1)*(1-((y(1)+yi(2))/K))-d*yi(3)*y(1);.........................1
vi(2) = r*yi(1)*((y(1)+yi(2)/K))-delta*yi(3)*y(2)-delta_m*yi(2);...............2
vi(3) = c*(((yilag1(1)+ylag1(2))/C)^x)*(1-yi(3));.......................................3
I want to solve this system of delay differential equations by dde23 solver....
where v i (1) = dvi/dt. (here i is in subcript)
v i (2) = dmi/dt
v i (3) = dai/dt
Somkenechukwu Mamah
Somkenechukwu Mamah 2021-10-25
Use
for i=0:6
Your starting value of i must start from 0 since your have (i-1) in your equation.
Just start your i value from 0 instaed of 1. This should fix the error.

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更多回答（4 个）

Jan
Jan 2019-3-11

0 个投票

Simplify your code:
ii = 4 * (i-1) + 1;
dy(ii) = r * y(ii) * (1 - (y(ii) + y(ii+1)) / K) - d * y(ii+2) * y(ii);
Now use the debugger:
dbstop if error
When Matlab stops at the error, check the values of the indices:
size(y)
ii
By the way, 5*10^(-1) is an expensive power operation and a multiplication, while 5e-1 or 0.5 is a cehap constant.

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parag gupta
parag gupta 2019-3-11
>> size(y)
ii
ans =
4 1
ii =
4
got these values ...I have created a zero matrix of order 4x1 ..( dy = zeros(4,1);) but still getting same error..
Jan
Jan 2019-3-11
If ii is 4 and y is a [4x1] vector, y(ii+1) and y(ii+2) must fail.
parag gupta
parag gupta 2019-3-11
or i =1
v = zeros(4,1);
v(1) = r*y(1)*(1-((y(1)+y(2))/K))-d*y(3)*y(1);
v(2) = r*y(1)*((y(1)+y(2)/K))-delta*y(3)*y(2)-delta_m*y(2);
v(3) = c*(((ylag1(1)+ylag1(2))/C)^x)*(1-y(3));
Actually these are 3 differential equations for one variant i.e v1,m1,a1 where dv1/dt = v(1) ,dm1/dt = v(2) ,da1/dt = v(2) and I solved them by using dde23 solver.
But now I want to find a solution of delay differential equations for i = 1: 6
vi(1) = r*yi(1)*(1-((y(1)+yi(2))/K))-d*yi(3)*y(1);.........................1
vi(2) = r*yi(1)*((y(1)+yi(2)/K))-delta*yi(3)*y(2)-delta_m*yi(2);...............2
vi(3) = c*(((yilag1(1)+ylag1(2))/C)^x)*(1-yi(3));.......................................3
where i is changing from 1 to 6 with different intial condition for different i.
so i wrote above code in which i got error.Could you please tell me how to edit my code so that i can get solution to system of differential equation which i have highlighted above..
thanks

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0 个投票

costo=abs((picos(I+1))-pico_ref)/pico_ref; %devuelve el pico siguiente
El índice supera el número de elementos de la matriz. El índice no debe exceder de 1.

0 个投票

Hello,
This code return the error message "Index exceeds the number of array elements. Index must not exceed 6.". The idea is read the values of table Excel and concatenate the values for every iterate of loop. What can I for correct this error and this problem.
Thanks
for ifile=1:nfiles
disp(['Traitement du fichier n° ',sprintf('%d',ifile)])
filename = fullfile('C:\Users\Stagiaire\Desktop\Enguerrand\Analyse_Pose_Docking',filelist(ifile).name); % Récupération du nom de fichier complet
disp(filename)
% Importation des données de RMSD et De Score de chaque fichier excel
% du répertoire de données
RMSD = xlsread(filename,'B:B');
Glide_Score = xlsread(filename,'C:C');
RMSD_Random = [RMSD(ifile); RMSD];
Glide_Score_Random = [Glide_Score(ifile); Glide_Score];
Analyse_Vec = horzcat(RMSD_Random, Glide_Score_Random); %Concatène les 2 vecteurs dans une matrice
disp(RMSD)
disp(Glide_Score)
disp(Glide_Score_Random)
disp(RMSD_Random)
end

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