Does logical indexing have a direct inverse?

Say I have some region-of-interest filter X, and I use B = A(X~=0) to only extract the values of A where X~=0, is it then possible to retrieve A from B and X?

I'm trying to think of a clever way to do it but coming up a little short!

Thanks for any suggestions,

Mike

[Martin Herrerias Azcue]

I come accross the same problem often enough that I wrote a function for it. It's really a matter of OCD, but I find:

A = revertfilter(B,f);

cleaner than:

A = zeros(size(B),'like',B);
A(f) = B;

Hope it helps.

function A = revertfilter(B,filter,dim,missing)
% A = REVERTFILTER(B,FILTER,DIM,MISSING) - Revert logical-indexing of A by FILTER upon dimension
%   DIM, filling any ~FILTER slices with MISSING.
%
%   E.g. if B = C(:,:,FILTER,:), A = REVERTFILTER(B,FILTER,3,NaN) will return a size(C) array,
%   such that all C(:,:,FILTER,:) == A(:,:,FILTER,:), and for which all A(:,:,~FILTER,:) == NaN.
%
% A = REVERTFILTER(B,FILTER,[],MISSING) - Revert logical-indexing B = A(FILTER), setting the size
%   of A from FILTER. For this to work DIM must be missing/empty, and nnz(FILTER) == numel(B).
%
%   E.g. C = rand(1,2,3); FILTER = C > 0.5; B = C(FILTER); A = REVERTFILTER(B,FILTER)

% (c) copyleft, Martin Herrerías 2019

    if nargin < 3, dim = []; end
    if nargin < 4, missing = 0; end
    
    assert(islogical(filter),'REVERSEFILTER only works for logical indexing');
    if isnumeric(B), missing = double(missing); 
    elseif islogical(B), missing = missing > 0;
    else
      assert(isequal(class(B),class(missing),'Class of MISSING must match B'));
    end

    if nnz(filter) == numel(B) && isempty(dim)
        A = emptyarray(size(filter),missing,B(1));
        A(filter) = B;
    elseif isvector(filter)
        if isempty(dim)
            if size(filter,2) > 1, dim = 2; else, dim = 1; end 
        end 
        assert(size(B,dim) == nnz(filter),'nnz(FILTER) ~= size(B,DIM)');
        sz = size(B);
        sz(dim) = numel(filter);
        A = emptyarray(sz,missing,B(1));
        
        % Place enough colons before and after dim for A(..,filter,..) = B to work
        args = [repmat({':'},1,dim-1),{filter},repmat({':'},1,numel(sz)-dim)];
        A(args{:}) = B;
    else
       error('Unrecognized syntax/arguments');
    end

end

function A = emptyarray(sz,missing,b)
% Not sure if this is worth the trouble: it might just be faster to use repmat(missing,sz),
% the idea is that the type of A is still set, whenever possible, by B

   if isnan(missing), A = NaN(sz,'like',b); return; end

   switch missing
       case 0, A = zeros(sz,'like',b);
       case 1, A = ones(sz,'like',b);
       case true, A = true(sz);
       case false, A = false(sz);
       case Inf, A = Inf(sz,'like',b);
       case -Inf, A = -Inf(sz,'like',b);
       otherwise, A = zeros(sz,'like',b); A(:) = missing;
   end
end

Answer 1

[Sean de Wolski]

I.e:

A(X~=0) = B

?

[Michael]

That looks the same as my forward problem, I'm looking for A = [matlab operation](B).

[Sean de Wolski]

No, the location data is required and it comes from X.

Also, you will never know about the values in A that were not extracted from X or the size of A. Thus you need the index to go in reverse.

Answer 2

[Walter Roberson]

No, it is not possible.

Suppose for example X was a matrix the same size as A, and is 1 for the top-left quadrant and 0 elsewhere. B = A(X~=0) would then retrieve values from the top left quadrant and would have no information about anything elsewhere in A. X also has no information about anything in A. Therefore, if you have only X and B, you cannot reconstruct the information that was in the other three quadrants of A.

My guess about what you actually want to do is:

A(X ~= 0) = B;

Answer 3

[Matt Fig]

Simple example:

A = randperm(4);
X = [0 1 0 1]
B = A(X~=0)
clear A

Now just look at X and B. By looking at just those two arrays, it is not even in principle possible to tell exactly what A was! This despite the fact that we know all of the elments of A! In a more general case, the problem is only worse. If you, as a human being, cannot tell what the original A was with such a simple example, how in the world could MATLAB tell?

Answer 4

[Image Analyst]

Michael: As the others have said, you can't get back ALL of a from only the part you extracted - you'd need the complete a to do that. However you extracted a chunk of "a" into a 1D column vector, and it is possible to put that back into a 2D reconstructed "a" where the extracted part is the same, but the unextracted part is zero (or any other number you want). Here's some demo code. See if it's what you were thinking.

% Generate sample data.
a = magic(8)
% Make x a mask in the upper left quadrant.
x = false(size(a)); % Initialize.
x(1:4, 1:4) = true % Mask of upper left quadrant
% Extract masked a into b.
b = a(x~=0) % This will be a 16 by 1 column vector.

% We can't get back ALL of a, without using a, but we CAN get back the a that we extracted
% back into the original 2D shape with zeros (or any constant) elsewhere.
a_reconstructed = zeros(size(x)); % Initialize a new "a" with 2D matrix of 0's.
% Find the linear indices that we used to extract elements of a.
linearIndexes = find(x~=0)
% Use them to assign those elements of a reconstructed a
% from the elements in b.
a_reconstructed(linearIndexes) = b(:)

Answer 5

[Chibuzo]

I believe the best approach will be to use set difference function, setdiff(A, B). For example: Given some indices of a vector A, you want to select other entries of A not included in the given indices.

A = randi([1, 15], 1, 10)

A = 1×10

15 9 6 10 10 14 2 6 2 13

indices = [2, 4, 7, 1]

indices = 1×4

2 4 7 1

A1 = A(indices)

A1 = 1×4

9 10 2 15

A2 = A(setdiff(1:length(A), indices))

A2 = 1×6

6 10 14 6 2 13

[DGM]

This doesn't really have much to do with the question. To simplify:

• We had an array A, but we don't necessarily know anything about it.
• We have a vector B containing some of the elements from A
• We have a logical mask X which is the same size as A

Knowing only B and X, reconstruct A.

Since we don't have A, the best we could do is insert the values in B into the appropriate locations in a new array (e.g. filled with zero or some padding values). There's no point in using setdiff(), since we're already using a logical array for indexing. Inverting the selection would be trivial, but it's not necessary for the assignment operation.

[Chibuzo]

Ok. Have you checked the sparse() and full() function? Sparse() removes 0's and squeezes. And full() returns the full matrix given a sparse matrix. You can condition you matrix accordingly to make this work.

[Rik]

Changing between sparse and full is a completely different operation, since the underlying mathematical object doesn't change, just how it is stored in memory. That is why it is possible to go back and forth between the two without issue.

This problem however, is completely different.

There really only is a solution in the trivial case that X contains only true, or in the case that all arrays are empty.