Is it possible to vectorize this?

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Omar Ali Muhammed
Omar Ali Muhammed 2021-4-27
编辑： Jan 2021-4-28
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 个评论
显示 4更早的评论
Matt J
Matt J 2021-4-27
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
Scott MacKenzie 2021-4-27
编辑：Scott MacKenzie 2021-4-27
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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采纳的回答

Sean de Wolski
Sean de Wolski 2021-4-27
编辑：Sean de Wolski 2021-4-27
Ran in:
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 个评论
Omar Ali Muhammed
Omar Ali Muhammed 2021-4-27
编辑：Omar Ali Muhammed 2021-4-27
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
Jan 2021-4-27
@Omar Ali Muhammed: Then move it to a function and provide A.' as input.

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更多回答（2 个）

Bruno Luong
Bruno Luong 2021-4-27
编辑：Bruno Luong 2021-4-27
Ran in:
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 个评论
Bruno Luong
Bruno Luong 2021-4-28
编辑：Bruno Luong 2021-4-28
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
Jan 2021-4-27
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 个评论
Bruno Luong
Bruno Luong 2021-4-27
I test for big matrix 1000 x 1000 and Jan's method is fatest.
Jan
Jan 2021-4-28
编辑：Jan 2021-4-28
@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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