Index exceeds the number of array elements .
541 ビュー (過去 30 日間)
古いコメントを表示
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?
1 件のコメント
Busuyi Adebayo
2022 年 2 月 22 日
Hi Gupta. Curious what process led to this type of model formulation in delay diffenetial equations. Can you share the process?
Thanks!
採用された回答
Guillaume
2019 年 3 月 11 日
編集済み: Guillaume
2019 年 3 月 11 日
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.
4 件のコメント
Somkenechukwu Mamah
2021 年 10 月 25 日
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.
その他の回答 (4 件)
Jan
2019 年 3 月 11 日
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.
3 件のコメント
Giomara Llumiquinga
2022 年 5 月 31 日
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 件のコメント
Enguerrand Galmiche
2023 年 2 月 13 日
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
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Matrix Indexing についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!