Index exceeds matrix dimensions.

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.

[the cyclist]

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?

Answer 1

[Image Analyst]

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.

Answer 2

[Walter Roberson]

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.

[Light]

FAILED AGAIN ;-((((

 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?

[Image Analyst]

I thought I answered it. You accepted my answer.

[Light]

I was wrong. It is not working

[Walter Roberson]

Here, index is the string '35', not the numeric value 35.

[Image Analyst]

See my answer to your new explanation in your duplicate question: http://www.mathworks.com/matlabcentral/answers/78455#answer_88183