我有一个矩阵如下：
id value
=============
1 0.5
2 0.5
3 0.8
4 0.3
5 0.2
从这个数组中，我希望找到总和小于或等于 1 的所有可能的组合。也就是说，
result
======
1 2
1 4 5
2 4 5
3 5
1 5
1 4
2 4
2 5
...
为了获得上述结果，我的想法是首先计算查找数组中元素之和的所有可能性，如下所示：
for ii = 1 : length(a) % compute number of possibilities
no_of_possibilities = no_of_possibilities + nchoosek(length(a),ii);
end
完成此操作后，然后循环遍历所有可能的组合。
我想知道是否有更简单的方法可以做到这一点。
data = [0.5, 0.5, 0.8, 0.3, 0.2];
required = cell(1, length(data));
subsets = cell(1, length(data));
for k = 2:length(data)-1 % removes trivial cases (all numbers or one number at a time)
% generate all possible k-pairs (if k = 3, then all possible triplets
% will be generated)
combination = nchoosek(1:length(data), k);
% for every triplet generated, this function sums the corresponding
% values and then decides whether then sum is less than equal to 1 or
% not
findRequired = @(x) sum(data(1, combination(x, :))) <= 1;
% generate a logical vector for all possible combinations like [0 1 0]
% which denotes that the 2nd combination satisfies the condition while
% the others do not
required{k} = arrayfun(findRequired, 1:size(combination, 1));
% access the corresponding combinations from the entire set
subsets{k} = combination(required{k}, :);
end
这会产生以下子集：
1 2
1 4
1 5
2 4
2 5
3 5
4 5
1 4 5
2 4 5
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