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alisha ali
alisha ali

Reputation: 21

Get elements of a matrix that are greater than sum of their two indices in row major order

I'm Writing a function called large_elements that takes input an array named X that is a matrix or a vector. The function identifies those elements of X that are greater than the sum of their two indexes.

For example, if the element X(2,3) is 6, then that element would be identified because 6 > (2 + 3). The output of the function gives the indexes(row and column sub) of such elements found in row-major order. It is a matrix with exactly two columns. The first column contains the row indexes, while the second column contains the corresponding column indexes.

Here is an example, the statement 

indexes = large_elements([1 4; 5 2; 6 0])

should give the output like this: 

[1 2; 2 1; 3 1]

If no such element exists, the function returns an empty array.

I have came up with the following code

function indexes = large_elements(A)
    [r c] = size(A);
    ind = 1;
    for ii = 1:r
        for jj = 1:c
            if A(ii,jj) > ii + jj
                indexes(ind,:) = [ii jj];
                ind = ind + 1;
            else
                indexes = [];
            end
       end  
    end
end

But the results are not as expected. Any help would be appreciated.

Upvotes: 1

Views: 921

Answers (2)

Santhan Salai
Santhan Salai

Reputation: 3898

One vectorised approch using bsxfun, find and ind2sub

A = randi(8,5);    %// Your matrix

%// finding sum of the indexes for all elements
indSum = bsxfun(@plus, (1:size(A,1)).', 1:size(A,2));

%// generating a mask of which elements satisfies the given condition (i.e A > indSum)
%// Transposing the mask and finding corresponding indexes
[c,r] = find(bsxfun(@gt, A, indSum).') ;

%// getting the matrix by appending row subs and col subs
out = [r,c]

Results:

Input A:

>> A

A =

 4     4     7     2     2
 1     3     4     8     3
 8     8     2     8     7
 8     3     4     5     1
 4     1     1     1     1

Output in row-major order:

out =

 1     1
 1     2
 1     3
 2     4
 3     1
 3     2
 3     4
 4     1

Note: Getting subs in row-major order is tricky here

Also here is your correct loopy approach

[r, c] = size(A);
ind = 0;
indexes = [];
for ii = 1:r
    for jj = 1:c
        if A(ii,jj) > ii + jj
            ind = ind + 1;
            indexes(ind,:) = [ii jj];              
        end
   end  
end

Upvotes: 0

Anindya Dutta
Anindya Dutta

Reputation: 1982

That is because whenever you encounter an element which is smaller than the sum of its indices you are reinitializing the array to null. So the output is coming out to be null. You should not initialize it to null on the else condition.

Upvotes: 0

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