我想提出另一个想法,它在概念上与erfan's。我的想法使用散列函数,具体来说,GetMD5FEX提交.
主要任务是如何“减少”中的每一行A
到单个代表值(例如字符向量),然后找到该向量的唯一条目。
Judging by the benchmark vs. the other suggestions, my answer doesn't perform as well as one of the alternatives, but I think its raison d'être lies in the fact that it is completely data-type agnostic (within the limitations of the GetMD5
1), that the algorithm is very straightforward to understand, it's a drop-in replacement as it operates on A
, and that the resulting array is exactly equal to the one obtained by the original method. Of course this requires a compiler to get working and has a risk of hash collisions (which might affect the result in VERY VERY rare cases).
以下是在我的计算机上典型运行的结果,后面是代码:
Original method timing: 8.764601s
Dev-iL's method timing: 0.053672s
erfan's method timing: 0.481716s
rahnema1's method timing: 0.009771s
function q39955559
G=[0 -1 1; 0 -1 2; 0 -1 3; 0 -1 4; 0 -1 5; 1 -1 6; 1 0 6; 1 1 6; 2 -1 6; 2 0 6; 2 1 6; 3 -1 6; 3 0 6; 3 1 6];
h=7;
M=reshape(G(nchoosek(1:size(G,1),h),:),[],h,size(G,2));
A=cell(size(M,1),2);
for p=1:size(M,1)
A{p,1}=squeeze(M(p,:,:));
left=~ismember(G, A{p,1}, 'rows');
A{p,2}=G(left,:);
end
%% Benchmark:
tic
A1 = orig_sort(A);
fprintf(1,'Original method timing:\t\t%fs\n',toc);
tic
A2 = hash_sort(A);
fprintf(1,'Dev-iL''s method timing:\t\t%fs\n',toc);
tic
A3 = erfan_sort(A);
fprintf(1,'erfan''s method timing:\t\t%fs\n',toc);
tic
A4 = rahnema1_sort(G,h);
fprintf(1,'rahnema1''s method timing:\t%fs\n',toc);
assert(isequal(A1,A2))
assert(isequal(A1,A3))
assert(isequal(numel(A1),numel(A4))) % This is the best test I could come up with...
function out = hash_sort(A)
% Hash the contents:
A_hashed = cellfun(@GetMD5,A,'UniformOutput',false);
% Sort hashes of each row:
A_hashed_sorted = A_hashed;
for ind1 = 1:size(A_hashed,1)
A_hashed_sorted(ind1,:) = sort(A_hashed(ind1,:));
end
A_hashed_sorted = cellstr(cell2mat(A_hashed_sorted));
% Find unique rows:
[~,ia,~] = unique(A_hashed_sorted,'stable');
% Extract relevant rows of A:
out = A(ia,:);
function A = orig_sort(A)
%To find equivalent rows up to order I use a double loop (VERY slow).
indices=[];
for j=1:size(A,1)
if ismember(j,indices)==0 %if we have not already identified j as a duplicate
for i=1:size(A,1)
if i~=j
if (isequal(A{j,1},A{i,1}) || isequal(A{j,1},A{i,2}))...
&&...
(isequal(A{j,2},A{i,1}) || isequal(A{j,2},A{i,2}))...
indices=[indices;i];
end
end
end
end
end
A(indices,:)=[];
function C = erfan_sort(A)
STR = cellfun(@(x) num2str((x(:)).'), A, 'UniformOutput', false);
[~, ~, id] = unique(STR);
IC = sort(reshape(id, [], size(STR, 2)), 2);
[~, col] = unique(IC, 'rows');
C = A(sort(col), :); % 'sort' makes the outputs exactly the same.
function A1 = rahnema1_sort(G,h)
idx = nchoosek(1:size(G,1),h);
%concatenate complements
M = [G(idx(1:size(idx,1)/2,:),:), G(idx(end:-1:size(idx,1)/2+1,:),:)];
%convert to cell so A1 is unique rows of A
A1 = mat2cell(M,repmat(h,size(idx,1)/2,1),repmat(size(G,2),2,1));
1 - If more complicated data types need to be hashed, one can use the DataHash FEX submission instead, which is somewhat slower.