Index exceeds matrix dimensions.
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
f = figure('Position',[10 10 600 600]);
dat = {-1,1,1;0,-1,0;0,0,-1;1,0,0};
cnames = {'1','2','5'};
rnames = {'1','2','3','4'};
t = uitable('Parent',f,'Data',dat,'ColumnName',cnames,...
'RowName',rnames,'Position',[10 10 590 590]);
blnA = logical( A == -1 );
blnOut = find(any(A == -1,2));
rnames{max(blnOut)};
negcolumn = find(A(min(blnOut),:) == 1);
cnames{max(negcolumn)};
dU(5)=7;
dU(cnames{max(negcolumn)})
Error message occurred. I couldn't find the mistake.(cnames{max(negcolumn)})=5 So it must be dU(5)=7 but error message
Index exceeds matrix dimensions.
1 commentaire
the cyclist
le 8 Juin 2013
I cannot run your code, because I do not have the array A. Is it possible for you to post self-contained code that will execute and show that error?
Réponse acceptée
Image Analyst
le 8 Juin 2013
Perhaps
A = cell2mat(dat);
but I can't figure out what dU is for - it's not being used. Anyway, look at this code:
dU(5)=7;
maxValue = max(negcolumn)
index = cnames{maxValue} % Gives string '5'
dU(5) % Works fine.
dU(cnames{max(negcolumn)}) % dU('5') = du(53) which is not defined because dU has only 5 elements.
You can't have the string '5' be the index into an array. Perhaps you wanted cnames to be a cell array that stored integers, not strings:
cnames = {1,2,5};
which works, though I still don't know what you're after.
0 commentaires
Plus de réponses (1)
Walter Roberson
le 8 Juin 2013
You have not defined dU()
cnames indexed by a number gives you a string. Indexing an array by a string is not impossible but seldom gives the expected answer: in your case it would be equivalent to asking for dU([49 50])
You need to remember that '1','2','5' are strings and that you almost never index arrays at strings; arrays get indexed at row numbers and column numbers.
If you need to be able to access an array by the name of a row, or the name of a column, then you need to use the MATLAB database object, which is part of the Statistics toolbox.
5 commentaires
Image Analyst
le 8 Juin 2013
See my answer to your new explanation in your duplicate question: http://www.mathworks.com/matlabcentral/answers/78455#answer_88183
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!