Matlab error for "Assignment has more non-singleton rhs dimensions than non-singleton subscripts".

Hi, i have error of "Assignment has more non-singleton rhs dimensions than non-singleton subscripts".

Error in Untitled10 (line 205)
            distAllTier1(i,j)= sqrt((node(i).x - node(j).x).^2 + (node(i).y - node(j).y).^2);

The details code as follow:

 for i = 1:1:numNodes
      for j = indPCHTier1

      if node(i).tier == 1
          if i == j
              distAllTier1(i,j) = NaN;
          else
              distAllTier1(i,j)= sqrt((node(i).x - node(j).x).^2 + (node(i).y - node(j).y).^2);
          end  
      else
          distAllTier1(i,j) = NaN;
          break;
        end
      end
  end

The result for

sqrt((node(i).x - node(j).x).^2 + (node(i).y - node(j).y).^2);

is:

ans =

    10.295630140987001
    21.213203435596427

Then the "distAllTier1(i,j)" cannot support the value. How can i solve this problem?

Thanks :)

Answer 1

[KSSV]

sqrt((node(i).x - node(j).x).^2 + (node(i).y - node(j).y).^2);

The output of above is 2*1..and you are trying to save output into a single scalar..

   distAllTier1(i,j)

so the error is popping out. I think the output of sqrt() should be a scalar. Show us the complete code to get more help.

[Asyran Abdullah]

Yes, correct. How can i change the input "distAllTier1(i,j)" into multi scalar?

I try to make like this, distAllTier1(i,j:2) but it doest not work.

The complete code is almost 500++ line, this problem came out because i change the matrix as my previous question in --> here

[Asyran Abdullah]

I attach again the related code:

numPCHTier1 = 0; 
for i = 1:1:numNodes 

      if node(i).pch == 1 && node(i).tier == 1 
          numPCHTier1 = numPCHTier1 + 1; 
          indPCHTier1(1,numPCHTier1) = i; 
      end 

end 

distAllTier1 = zeros(numNodes,numPCHTier1); 
for i = 1:1:numNodes
      for j = indPCHTier1
      if node(i).tier == 1
          if i == j
              distAllTier1(i,j) = NaN;
          else
              distAllTier1(i,j)= sqrt((node(i).x - node(j).x).^2 + (node(i).y - node(j).y).^2);
          end  
      else
          distAllTier1(i,j) = NaN;
          break;
        end
      end
  end

[KSSV]

Try using cells...

distAllTier1 = cell(numNodes,indPCHTier1) ;
count = 0 ;
for i = 1:1:numNodes
    for j = indPCHTier1
        count = count+1 ;
        if node(i).tier == 1           
            if i == j
                distAllTier1{i,j} = NaN;
            else
                distAllTier1{i,j}= sqrt((node(i).x - node(j).x).^2 + (node(i).y - node(j).y).^2);
            end
        else
            distAllTier1{i,j} = NaN;
            break;
        end
    end
end

[Asyran Abdullah]

Thanks @KSSV for your help, when change to cell it appear error like this:

Error using cell
Size inputs must be scalar.

[KSSV]

There was a typo error..now modified...check it now.

[Asyran Abdullah]

WOW! you are so fantastic! Thanks it works!

However, i change the line to this:

distAllTier1 = cell(numNodes,numPCHTier1) ;

Thanks

[Asyran Abdullah]

I'm sorry sir, regarding for changing into cell, i got an issues here :

distAllTier1(distAllTier1 == 0) = NaN;

Above line appear some error:

Undefined operator '==' for input arguments of type 'cell'.

I try to change ( ) to { }, its still same.

Thanks

[KSSV]

You cannot use cells like that. See this:

k{1} = [1 0] ;
idx = k{1}==0 ;
k{1}(idx) = NaN

[Asyran Abdullah]

Then i change k into distAllTier1. Thanks sir!

[Asyran Abdullah]

Sir, one more sir.

for i = 1:1:noTier1

          if node(i).pch ~= 1
              [minVal,index] = min(distAllTier1(i,:));
              node(i).head = index;
          end
      end

It contain error:

Undefined function 'min' for input arguments of type 'cell'.

Error in Untitled10 (line 267)
            [minVal,index] = min(distAllTier1(i,:));

The cell cannot read min value. Again i try to use { } but Matlab said,

Error using min
Too many input arguments.

I hope u can help me for last time. Thanks Dr.

[KSSV]

min([distAllTier1{i,:}]);

[Asyran Abdullah]

I try to find the solution, but could not solve it. After do min([distAllTier1{i,:}]); it appear this error:

Error using horzcat
Dimensions of matrices being concatenated are not consistent.

    Error in Untitled10 (line 267)
                [minVal,index] = min([distAllTier1{i,:} ] );

[KSSV]

[distAllTier1{i,:}]

Try the above line alone.....this should give you a vector.

[Asyran Abdullah]

Thanks sir, i appreciate your time and help .