这是我想出来的
test = [1,2,3;4,5,6;7,8,9]; % orig matrix
[m, n] = size(test); % example 1, breaks with value zero elems
o = find(test);
test1 = [o, reshape(test, m*n, 1), o]
经过的时间是 0.004104 秒。
% one liner from above
% (depending on data size might want to avoid dual find calls)
test2=[ find(test) reshape(test, size(test,1)*size(test,2), 1 ) find(test)]
经过的时间是 0.008121 秒。
[r, c, v] = find(test); % just another way to write above, still breaks on zeros
test3 = [r, v, c]
经过的时间是 0.009516 秒。
[i, j] =ind2sub([m n],[1:m*n]); % use ind2sub to build tables of indicies
% and reshape to build col vector
test4 = [i', reshape(test, m*n, 1), j']
经过的时间是 0.011579 秒。
test0 = [1,2,3;0,5,6;0,8,9]; % testing find with zeros.....breaks
% test5=[ find(test0) reshape(test0, size(test0,1)*size(test0,2), 1 ) find(test0)] % error in horzcat
[i, j] =ind2sub([m n],[1:m*n]); % testing ind2sub with zeros.... winner
test6 = [i', reshape(test0, m*n, 1), j']
经过的时间是 0.014166 秒。
从上面使用网格网格:
经过的时间是 0.048007 秒。