What is the error, attempt to grow an array along ambiguous dimension?

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Billy 2012-12-3

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Hello, what is the error attempt to grow an array along ambiguous dimension? And, how would I be able to fix that. The code that gives the error is.

    for j = 1:rep
    canv2 = canv;
    del = j*inc;
    canv2(1+del:row+del,1+del:col+del,:) = cm;
    filter = find(canv2 > 0);
    canv3 = bg; 
    canv3(filter) = canv2(filter); 
    image(canv3)
    pause(0.05)
end

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Billy 2012-12-3

The error apparently involves line of canv3(filter)=canv2(filter); and I tried changing that line of code and the error still came up. I'll try checking the size.

Jan 2012-12-3

@Billy: Please do not only mention, that you have changed the code, but show us the new code also. We do not have crystal balls (except for Walter, of course).

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Answer 1

Jan 2012-12-3

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When canv3 is an array with more than 1 non-singelton dimension (e.g. 2D matrix or 3D), using a linear or logical index for growing cannot be interpreted uniquely anymore:

x = [1,2; 3,4];
x([true, true, true, true, true, true]) = 7;

Does this create the same error message? If so, Matlab cannot decide, if the new array should be [2x3] or [3x2].

Avoid using the name "filter" for variables, because it is a built-in function
Let canv3 grow explictly at first:

index = (canv2 > 0);  % not "filter"
s = size(canv2);
if any(size(canv3) < s)
  canv3(s(1), s(2), s(3)) = 0;  % Grow
end
canv3(index) = canv2(index);

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Billy 2012-12-4

编辑：Billy 2012-12-4

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I get the error where canv3(index)=canv2(index).

 Attempt to grow array along ambiguous dimension.
 Error in imageTestTranslate (line 54)
    canv3(index) = canv2(index);

Matt J 2012-12-4

编辑：Matt J 2012-12-4

You cannot use linear indexing or other kinds of 1-dimensional indexing to grow an array that is not a vector.

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Answer 2

Matt J 2012-12-3

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编辑：Matt J 2012-12-3

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Copy/paste the error messages, so we can see what lines generate them and also the output of WHOS so we can see the sizes of all your variables.

My guess would be that it occurs in this line

canv3(filter) = canv2(filter);

and that it occurs because "filter" contains indices greater than numel(canv3).

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Billy 2012-12-4

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Yes, that appears to be true. How could I make the number of elements the same?

numel(filter)
ans =
     3959856
 numel(canv3)
 ans =
     3932160

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Answer 3

Matt J 2012-12-4

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编辑：Matt J 2012-12-4

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If the whole point of all this is to translate an image by an integer amount, just use circshift

canv3=circshift(canv2,shiftvector) + bg;

or the function below if you want non-circulant shifting.

    function [B,src_indices,dest_indices]=noncircshift(A,offsets)
    %Like circshift, but shifts are not circulant. Missing data are filled with
    %zeros.
    %
    %  [B,src_indices,dest_indices]=noncirchift(A,offsets)
    %
    %B is the resulting array and the other outputs are such that
    %
    %  B(dest_indices{:})=A(src_indices{:})
    siz=size(A);
    N=length(siz);
    if length(offsets)<N
       offsets(N)=0; 
    end
    B=zeros(siz);
    indices=cell(3,N);
    for ii=1:N
        for ss=[1,3]
         idx=(1:siz(ii))+(ss-2)*offsets(ii);
          idx(idx<1)=[];
          idx(idx>siz(ii))=[];
         indices{ss,ii}=idx;
        end
    end
    src_indices=indices(1,:);
    dest_indices=indices(3,:);
    B(dest_indices{:})=A(src_indices{:});