Index exceeds matrix dimensions.AGAIN

FAILED AGAIN ;-(((( Accepted wrong one
dU(35)=55;
f = figure('Position',[10 10 600 600]);
dat = {-1,1,1;0,-1,0;0,0,-1;1,0,0};
cnames = {'18','29','51'};
rnames = {'12','26','35','44'};
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));
maxValue = max(negcolumn)
index = rnames{maxValue}
dU(tzan)
maxValue =
3
index =
35
Index exceeds matrix dimensions.
Please give me easy way...
Help, Help, Help It is my last question?

6 件のコメント

Light
Light 2013 年 6 月 8 日
I don't need vector form. I just need regular number
it must be
dU(tzan)=55
But it is error
Index exceeds matrix dimensions.
Light
Light 2013 年 6 月 8 日
編集済み: Light 2013 年 6 月 8 日
dU(35)=59;
t=35;
dU(t)
ans =
59
t=index
t =
35
>> dU(t)
""""Index exceeds matrix dimensions.""""
Don't make me cry ;-(
Image Analyst
Image Analyst 2013 年 6 月 8 日
編集済み: Image Analyst 2013 年 6 月 8 日
Please stop posting the question over and over again. Learn how to use the debugger: http://blogs.mathworks.com/videos/2012/07/03/debugging-in-matlab/. It will solve all your problems. It really will. We can't help you because we don't know what A is. Put a "clear all" at the beginning of your code and get rid of that pre-existing A. You will find your code crashes when it tries to access A. Then proceed to fix your code using the debugger.
Light
Light 2013 年 6 月 8 日
Sorry i'm nervous. I will change it clear!
Image Analyst
Image Analyst 2013 年 6 月 8 日
No one even knows what you want to do. The code is so bad that we can't even figure it out. Please explain, in words, what you want to do with the cnames and rnames and the table. It seems like you're looking for -1's but I don't know what you want to do if you find them.
Azzi Abdelmalek
Azzi Abdelmalek 2013 年 6 月 8 日
Light, can you enlighten us by making your question breve and clear, or maybe you need to learn basics of Matlab

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