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Obtain every possible combination in array

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Tyler Murray
Tyler Murray on 30 Mar 2017
Commented: Walter Roberson on 8 May 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.


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David Goodmanson
David Goodmanson on 30 Mar 2017
Hi Tyler, As you said, you are talking one large matrix. Assuming that the choices from the first vector are two independent choices, then you will have a matrix with 100x100x100x100 = 1e8 rows and 4 columns. You may want to rethink this.
Also, if your vectors contain the usual Matlab real numbers at 8 bytes apiece to store, that's 3.2 gigabytes of required memory.
Tyler Murray
Tyler Murray on 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 on 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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Answers (1)

Kirby Fears
Kirby Fears on 30 Mar 2017
Edited: Kirby Fears on 8 May 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)];

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