Attempt to grow array along ambiguous dimension. it happens when N,M are smaller then the total of A,B. can I make my code work for any value of N,M? if so how would I do that

N=5;

M=4;

A=[1 2 5 9 ; 23 23 874 243; 5 4 6 7; 5 09 23 31];

B=[7 8; 9 10; 12 11];

C=sort([A(:); B(:)]);

res=zeros(N,M);

res(1:length(C(:)))=C(1:end)

%the error I get is:

Attempt to grow array along ambiguous dimension.

Error on res(1:length(C(:)))=C(1:end)

Answer 1

[Dyuman Joshi]

The number of elements in C is greater than the total number of elements pre-allocated in res.

You can not put 22 elements in 20 place holders, where there is only one element per place holder; no matter the arrangement.

Nor can you grow it along any of the size/dimension to accomodate the extra elements, as stated in the error.

N=5; M=4;
A=[1 2 5 9 ; 23 23 874 243; 5 4 6 7; 5 09 23 31];
B=[7 8; 9 10; 12 11];
C=sort([A(:); B(:)]);
numel(C)

ans = 22

res=zeros(N,M);
numel(res)

ans = 20

"can I make my code work for any value of N,M? if so how would I do that"

Yes, your code will work for a values of N and M, iff N*M>=22

%%Examples -
%N=5, M=5, N*M=25 which is grater than 22
res1=zeros(5,5);
res1(1:numel(C))=C

res1 = 5×5

1 5 9 12 243 2 6 9 23 874 4 7 9 23 0 5 7 10 23 0 5 8 11 31 0

%N=2,M=11, N*M=22 which is equalto 22
res2=zeros(2,11);
res2(1:numel(C))=C

res2 = 2×11

1 4 5 6 7 9 9 11 23 23 243 2 5 5 7 8 9 10 12 23 31 874

[Joshua]

Thank you, this helped a lot.

Answer 2

[Jan]

Some simplifications:

• Use numel(C) instead of length(C(:)).
• C(1:end) is exactly the same as C.

The values to not matter the problem. A shorter version, which explains the problem:

res = zeros(5, 4);
C   = ones(22, 1);
res(1:numel(C)) = C;

Attempt to grow array along ambiguous dimension.

Matlab cannot guess, what the shape of res should be after this code and I can't also. There is no unique decision how to expand a [5x4] matrix to contan 22 elements.

I cannot suggest a solution, because it is unclear, what you want to achieve. This is the meaning of the error message.

[Joshua]

Thank you so much for your help, I did not think of those simplifications!

Answer 3

[Image Analyst]

Try this, which handles both cases: where res has more elements than C and where res has fewer elements than C:

rows = 5;
columns = 4;
A = [1 2 5 9 ; 23 23 874 243; 5 4 6 7; 5 09 23 31];
B = [7 8; 9 10; 12 11];
C = sort([A(:); B(:)]);
res = zeros(rows, columns);
if numel(res) < numel(C)
    linearIndexes = 1 : numel(res);
else
    linearIndexes = 1 : numel(C);
end
res(linearIndexes) = C(linearIndexes)