# Is it possible to vectorize this?

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Omar Ali Muhammed on 27 Apr 2021
Edited: Jan on 28 Apr 2021
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....
##### 6 CommentsShowHide 5 older comments
Scott MacKenzie on 27 Apr 2021
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.

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### Accepted Answer

Sean de Wolski on 27 Apr 2021
Edited: Sean de Wolski on 27 Apr 2021
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
##### 2 CommentsShowHide 1 older comment
Jan on 27 Apr 2021
@Omar Ali Muhammed: Then move it to a function and provide A.' as input.

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### More Answers (2)

Bruno Luong on 27 Apr 2021
Edited: Bruno Luong on 27 Apr 2021
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} = []
##### 1 CommentShowHide None
Bruno Luong on 28 Apr 2021
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

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Jan on 27 Apr 2021
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.
##### 2 CommentsShowHide 1 older comment
Jan on 28 Apr 2021
@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.

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