problem with combvec, allcomb
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How to use combvec for a huge set (a1 to a100)?
Example a1=[1 3 4], a2=[5 7], a3=[7 9 1], .... a100=[8 9 6 2]
댓글 수: 5
Walter Roberson
2012년 11월 9일
The number of results you will get will be the product of the number of elements in the sets. That could be quite a lot!!
답변 (2개)
Andrei Bobrov
2012년 11월 10일
편집: Andrei Bobrov
2012년 11월 11일
a = {1,3,[2 6 4],[8 9]};
c = cell(size(a));
[c{:}] = ndgrid(a{end:-1:1});
c2 = sortrows(cell2mat(cellfun(@(x)x(:),c,'un',0)),1);
out = fliplr(c2)';
ADD
n = cellfun(@numel,a);
t = max(n)./n;
while any(rem(t,1) ~= 0)
t = t*2;
end
c = cellfun(@(x,y)repmat(x,1,y),a,num2cell(t),'un',0);
out = cat(1,c{:});
댓글 수: 3
Rik
2022년 3월 2일
combvec全组合
(machine translation of the Chinese characters: "full combination")
Matt Fig
2012년 11월 9일
This will solve your problem.
function M = expand_vects(A)
% A is a cell array of 1-by-n vectors where n need not
% be the same in every element.
L = cellfun('length',A);
B = lcms(L);
M = zeros(length(L),B);
for ii = 1:length(L)
M(ii,:) = reshape(A{ii}(ones(1,B/L(ii)),:).',1,B);
end
Once you have the files in place, call like this:
>> A = {1,3,[2,6,4],[8,9]}; % Use a cell array
>> M = expand_vects(A)
M =
1 1 1 1 1 1
3 3 3 3 3 3
2 6 4 2 6 4
8 9 8 9 8 9
댓글 수: 2
Matt Fig
2012년 11월 9일
You can forgo the FEX function and just use this. (I found that recursively calling LCM is just as fast or faster than the FEX function.)
function M = expand_vects(A)
% A is a cell array of 1-by-n vectors where n need not
% be the same in every element.
L = cellfun('length',A);
LC = length(L);
% B = lcms(L); % Available on the FEX. Or use below loop.
B = L(1);
for ii = 2:LC
B = lcm(B,L(ii));
end
M = zeros(LC,B);
L = B./L;
for ii = 1:LC
M(ii,:) = reshape(A{ii}(ones(1,L(ii)),:).',1,B);
end
Matt Fig
2012년 11월 10일
Well.....
Simply make yourself a cell array and pass it in:
a1 = 1;
a2 = 3;
a3 = [3 4 5];
A = {a1,a2,a3};
Nasty problems like this are the reason we don't make variables a1,a2,a3,a4.....an in the first place.....
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