Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
570 views
in Technique[技术] by (71.8m points)

matlab - Generate all possible combinations of the elements of some vectors (Cartesian product)

I would like to generate all the possible combinations of the elements of a given number of vectors.

For example, for [1 2], [1 2] and [4 5] I want to generate the elements:

[1 1 4; 1 1 5; 1 2 4; 1 2 5; 2 1 4; 2 1 5; 2 2 4; 2 2 5]

The problem is that I don't know the number of vectors for which I need to calculate the combinations. There might be 3 as in this case, or there may be 10, and I need a generalization. Can you please help me to this in MATLAB? Is there already a predefined function that can do this task?

Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Consider this solution using the NDGRID function:

sets = {[1 2], [1 2], [4 5]};
[x y z] = ndgrid(sets{:});
cartProd = [x(:) y(:) z(:)];

cartProd =
     1     1     4
     2     1     4
     1     2     4
     2     2     4
     1     1     5
     2     1     5
     1     2     5
     2     2     5

Or if you want a general solution for any number of sets (without having to create the variables manually), use this function definition:

function result = cartesianProduct(sets)
    c = cell(1, numel(sets));
    [c{:}] = ndgrid( sets{:} );
    result = cell2mat( cellfun(@(v)v(:), c, 'UniformOutput',false) );
end

Note that if you prefer, you can sort the results:

cartProd = sortrows(cartProd, 1:numel(sets));

Also, the code above does not check if the sets have no duplicate values (ex: {[1 1] [1 2] [4 5]}). Add this one line if you want to:

sets = cellfun(@unique, sets, 'UniformOutput',false);

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...