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

2018 9 27
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更新时间：2018 9 27
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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 :)

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 采纳的回答

KSSV
KSSV 2018-9-27

1 个投票

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.

14 个评论
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Asyran Abdullah
Asyran Abdullah 2018-9-27
编辑：Asyran Abdullah 2018-9-27
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
Asyran Abdullah 2018-9-27
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
KSSV 2018-9-27
编辑：KSSV 2018-9-27
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
Asyran Abdullah 2018-9-27
Thanks @KSSV for your help, when change to cell it appear error like this:
Error using cell
Size inputs must be scalar.
Asyran Abdullah
Asyran Abdullah 2018-9-27
编辑：Asyran Abdullah 2018-9-27
WOW! you are so fantastic! Thanks it works!
However, i change the line to this:
distAllTier1 = cell(numNodes,numPCHTier1) ;
Thanks
Asyran Abdullah
Asyran Abdullah 2018-9-27
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
KSSV 2018-9-27
You cannot use cells like that. See this:
k{1} = [1 0] ;
idx = k{1}==0 ;
k{1}(idx) = NaN
Asyran Abdullah
Asyran Abdullah 2018-9-27
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.
Asyran Abdullah
Asyran Abdullah 2018-9-27
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
KSSV 2018-9-27
[distAllTier1{i,:}]
Try the above line alone.....this should give you a vector.

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