Is it possible to vectorize this?

I want to find the most occurring element in the matrix in column wise excluding zeror elements.

e.g if A = [1 2 0 3 4 6; 9 3 4 0 9 5; 4 3 0 5 6 7; 3 7 7 3 0 0;1 1 8 8 4 8; 0 0 0 0 4 2; 0 0 0 0 0 0]'

The result is a cell matrix B

B={[1 2 3 4 6], [9], [4 3 5 6 7],[3 7], [8],[4 2], nan}

So most occurring elements is a cell array.

loops are inefficient for lage matrix.

Thanks in advance....

[Dyuman Joshi]

Shouldn't B be, B={[1 2 3 4 6], [9], [4 3 5 6 7],[3 7], [8]} ?

[Omar Ali Muhammed]

It Should be, B={[1 2 3 4 6], [9], [4 3 5 6 7],[3 7], [8],[4 2], nan}

Regards

[Dyuman Joshi]

My bad, Overlooked the Nan somehow. How large are the input matrix and what code have you tried yet? (most importantly the loop part)

[Matt J]

I want to find the most occurring element in the matrix in column wise excluding zeror elements.

Don't you mean row-wise?

[Matt J]

loops are inefficient for lage matrix.

Note that in Matlab, there is no way of populating a cell array that doesn't involve a for-loop (or something the same speed as a for-loop). So, the use of loops in some way will be inevitable.

[Scott MacKenzie]

Just to clarify, your question says "column wise". Do you mean "row wise"? Your example solution, B, shows the most occurring elements along the rows in A.

The code below avoids a loop and gets close to your goal:

% Assume A is the initial matrix (as in the example)
A(A==0) = NaN;
[~, ~, B] = mode(A,2);
B = B'

If A is your example matrix, then B matches your example result, except for the NaN entries where 0 occurs. Oddly (to me, anyway), you want 0s excluded except in the situation where all elments in a row are 0. In that case, NaN appears as the most occurring element. That doesn't quite add up to me, but that's the logic I see in your example.

Answer 1

[Sean de Wolski]

A = [1 2 0 3 4 6; 9 3 4 0 9 5; 4 3 0 5 6 7; 3 7 7 3 0 0;1 1 8 8 4 8; 0 0 0 0 4 2; 0 0 0 0 0 0]

A = 7×6

1 2 0 3 4 6 9 3 4 0 9 5 4 3 0 5 6 7 3 7 7 3 0 0 1 1 8 8 4 8 0 0 0 0 4 2 0 0 0 0 0 0

B = accumarray(repmat((1:height(A)).',width(A),1),A(:), [],@(x)modeall(nonzeros(x)))

B = 7×1 cell array
    {5×1 double}
    {[       9]}
    {5×1 double}
    {2×1 double}
    {[       8]}
    {2×1 double}
    {[     NaN]}

celldisp(B)

 
B{1} =
 
     1
     2
     3
     4
     6

 
 
B{2} =
 
     9

 
 
B{3} =
 
     3
     4
     5
     6
     7

 
 
B{4} =
 
     3
     7

 
 
B{5} =
 
     8

 
 
B{6} =
 
     2
     4

 
 
B{7} =
 
   NaN

 

function m = modeall(x)
    [~,~,m] = mode(x);
    if isempty(m{1}) % Handle empty case
        m{1} = nan;
    end
end

[Omar Ali Muhammed]

The code process the matrix row-wise, not column wise.

A =

1 9 4 3 1 0 0

2 3 3 7 1 0 0

0 4 0 7 8 0 0

3 0 5 3 8 0 0

4 9 6 0 4 4 0

6 5 7 0 8 2 0

[Jan]

@Omar Ali Muhammed: Then move it to a function and provide A.' as input.

Answer 2

[Bruno Luong]

NOTE the order of most is sorted with this algorithm:

A = [1 2 0 3 4 6; 
    9 3 4 0 9 5; 
    4 3 0 5 6 7; 
    3 7 7 3 0 0;
    1 1 8 8 4 8; 
    0 0 0 0 4 2; 
    0 0 0 0 0 0]'

A = 6×7

1 9 4 3 1 0 0 2 3 3 7 1 0 0 0 4 0 7 8 0 0 3 0 5 3 8 0 0 4 9 6 0 4 4 0 6 5 7 0 8 2 0

% Algo
[u,~,I] = unique(A);
keep = A ~= 0;
[~,J] = find(keep);
c = accumarray([I(keep),J],1);
[r,c] = find(c == max(c,[],1) & c>0);
B = accumarray(c,r,[size(A,2) 1], @(r) {u(r)})';

celldisp(B)

 
B{1} =
 
     1
     2
     3
     4
     6

 
 
B{2} =
 
     9

 
 
B{3} =
 
     3
     4
     5
     6
     7

 
 
B{4} =
 
     3
     7

 
 
B{5} =
 
     8

 
 
B{6} =
 
     2
     4

 
 
B{7} =
 
     []
 

[Bruno Luong]

In case A contains reasonably small integers, the UNIQUE command can be removed and this method can be faster

% I = A; % <= this replace UNIQUE
keep = A ~= 0;
[~,J] = find(keep);
c = accumarray([A(keep),J],1);
[r,c] = find(c == max(c,[],1) & c>0);
B = accumarray(c,r,[size(A,2) 1], @(r) {r})'; % indexing u{r} is no longer needed

Answer 3

[Jan]

mode() handles matrices as inputs also. Only ignoring the zeros is complicated.

For a comparison here the loop method:

A = [1 2 0 3 4 6; 9 3 4 0 9 5; 4 3 0 5 6 7; 3 7 7 3 0 0;1 1 8 8 4 8; 0 0 0 0 4 2; 0 0 0 0 0 0];
C = ModeFull(A.');
celldisp(C)

function C = ModeFull(A)
% Mode along 1st dimension ignoring zeros
n = size(A, 2);
C = cell(1, n);
for k = 1:n
   a = A(:, k);
   a = a(a ~= 0);
   if isempty(a)
       C{k} = NaN;
   else
       x     = sort(a);
       start = find([true; diff(x) ~= 0]);
       freq  = zeros(numel(x), 1);
       freq(start) = [diff(start); numel(x) + 1 - start(end)];
       m     = max(freq);
       C{k}  = x(freq == m).';
   end 
end
end

Please compare the run time with Sean de Wolski's vectorized approach for your real data.

[Bruno Luong]

I test for big matrix 1000 x 1000 and Jan's method is fatest.

[Jan]

@Bruno Luong: Some timings (i5 mobile, R2018b)

A = randi(50, 1000, 1000);
A(rand(size(A)) < 0.2) = 0;

tic
B = accumarray(repmat((1:size(A, 1)).', size(A, 2), 1), A(:), [], ...
   @(x)modeall(nonzeros(x)));
toc

tic; C = BrunosMode(A.'); toc

tic; D = ModeFull(A.'); toc

% Elapsed time is 0.402765 seconds.  Sean
% Elapsed time is 0.165996 seconds.  Bruno
% Elapsed time is 0.075373 seconds.  Jan

This is another example, where the assumption "loops are inefficient for large matrices" do not match the expectations. This was the case before the JIT become powerful in Matlab 6.5 - this was in 2002. But as the "brute clearing header" the rumor of slow loops is still living.

Vectorizing is very efficient, if the data and the operation is suitable and if no huge intermediate data are produced.