问题
If I have this vector:
x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6]
I would like to get the position of each unique number according to itself.
y = [1 2 3 4 5 1 2 3 1 1 2 1 2 3 4]
At the moment I'm using:
y = sum(triu(x==x.')) % MATLAB 2016b and above
It's compact but obviously not memory efficient.
For the pure beauty of MATLAB programming I would avoid using a loop. Do you have a better simple implementation ?
Context:
My final goal is to sort the vector x but with the constraint that a number that appear N times has the priority over another number that has appeared more than N times:
[~,ind] = sort(y);
x_relative_sort = x(ind);
% x_relative_sort = 1 2 3 4 6 1 2 4 6 1 2 6 1 6 1
回答1:
Assuming x is sorted, here's one vectorized alternative using unique, diff, and cumsum:
[~, index] = unique(x);
y = ones(size(x));
y(index(2:end)) = y(index(2:end))-diff(index).';
y = cumsum(y);
And now you can apply your final sorting:
>> [~, ind] = sort(y);
>> x_relative_sort = x(ind)
x_relative_sort =
1 2 3 4 6 1 2 4 6 1 2 6 1 6 1
回答2:
If you have positive integers you can use sparse matrix:
[y ,~] = find(sort(sparse(1:numel(x), x, true), 1, 'descend'));
Likewise x_relative_sort can directly be computed:
[x_relative_sort ,~] = find(sort(sparse(x ,1:numel(x),true), 2, 'descend'));
回答3:
Just for variety, here's a solution based on accumarray. It works for x sorted and containing positive integers, as in the question:
y = cell2mat(accumarray(x(:), x(:), [], @(t){1:numel(t)}).');
回答4:
You can be more memory efficient by only comparing to unique(x), so you don't have a large N*N matrix but rather N*M, where N=numel(x), M=numel(unique(x)).
I've used an anonymous function syntax to avoid declaring an intermediate matrix variable, needed as it's used twice - this can probably be improved.
f = @(X) sum(cumsum(X,2).*X); y = f(unique(x).'==x);
回答5:
Here's my solution that doesn't require sorting:
x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
y = cell2mat( splitapply(@(v){cumsum(v)},x,cumsum(logical([1 diff(x)]))) ) ./ x;
Explanation:
% Turn each group new into a unique number:
t1 = cumsum(logical([1 diff(x)]));
% x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
% t1 = [1 1 1 1 1 2 2 2 3 4 4 5 5 5 5 6 6 6];
% Apply cumsum separately to each group:
t2 = cell2mat( splitapply(@(v){cumsum(v)},x,t1) );
% t1 = [1 1 1 1 1 2 2 2 3 4 4 5 5 5 5 6 6 6];
% t2 = [1 2 3 4 5 2 4 6 3 4 8 6 12 18 24 1 2 3];
% Finally, divide by x to get the increasing values:
y = t2 ./ x;
% x = [1 1 1 1 1 2 2 2 3 4 4 6 6 6 6 1 1 1];
% t2 = [1 2 3 4 5 2 4 6 3 4 8 6 12 18 24 1 2 3];
来源：https://stackoverflow.com/questions/54079558/count-repeating-integers-in-an-array