Obtain every possible combination in array

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Tyler Murray
Tyler Murray il 30 Mar 2017
I have 3, 100 x 1 vectors that I am combining to make a m x 4 matrix. To obtain 4 columns in the desired output, two elements from the first vector are selected as the first two elements in a row, and then one element from the second vector as the third element, and one element from the third vector as the fourth element to complete the m x 4. I would like to get every possible row combination possible in one large matrix. I started with a huge nested for loop and quickly got lost and assume there must be a better way. Any help is appreciated.
  4 Commenti
Tyler Murray
Tyler Murray il 30 Mar 2017
By possible I mean row indices. Order does not matter I am just looking for the total number of combinations. Thus no permutations necessary. Thanks for your reply, I'd appreciate any help.
Kirby Fears
Kirby Fears il 30 Mar 2017
Modificato: Kirby Fears il 30 Mar 2017
The combinations total to (100)*99/2*100*100 = 49,500,000 rows. It's probably worth rethinking this approach. I will post some code below anyway...

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Risposte (1)

Kirby Fears
Kirby Fears il 30 Mar 2017
Modificato: Kirby Fears il 8 Mag 2017
Here's a brute force generation of all combinations. Memory usage is more of a bottleneck than processing time, so I wrote it with for-loops.
Note that storing all combinations may not be the best approach to address your underlying use case.
Set sz = 100 to match your vector size; beware of 50 million rows.
% Setting up dummy data
sz = 10; % 4500 rows
a = (1:sz)';
b = a + sz;
c = b + sz;
res = NaN(sz^3*(sz-1)/2,4);
% Filling array with selections
iCounter = 0;
for ai = 1:(numel(a)-1),
for aj = (ai+1):numel(a),
for bi = 1:sz,
for ci = 1:sz,
iCounter = iCounter + 1;
res(iCounter,:) = [a(ai) a(aj) b(bi) c(ci)];
end
end
end
end

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