How can I create a for loop when the range is changing

Hello,

I have a code as following. Code works as intended it sorts the elements as I want. After that I want to print the first element of sorted array and the corresponding distance. Actually each distance represents a song but I didn't write their names therefore I use the index of distance. My problem is there are some duplicates is sorted array. So, when I run the code if there are two times 0.5 distance it prints the same line 2 times. I somehow need to delete all coppies of sorted(i) without changing the lenght of it. If I just delete it since i goes until 19 sorted(i) will give error of index out of range. How can I solve this problem?

clc;clear;
Crescendo = [0 1 0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 0]';
Descrescendo = [1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 0 0]';
Staccato = [0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0]';
Portamento = [0 0 0 1 0 1 1 1 0 0 0 1 1 0 0 1 0 1 0 0]';
Tempo = [1 1 1 0 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1]';
Intervals = [0 1 1 1 1 1 0 1 1 0 0 0 1 0 0 1 1 1 1 1]';
M = [Crescendo Descrescendo Staccato Portamento Tempo Intervals];
S = M*M';
[U,W,V] = svds(S,2);
%labels = cellstr(num2str([1 : length(U)]'));
%plot(U(: ,1),U(: ,2),'rx');
%text(U(: ,1),U(: ,2), labels)
distance = zeros(19,1);
for i=2:20
    distance(i-1,1) = dot(U(1,:),U(i,:))/(norm(U(1,:))*norm(U(i,:)));
end
sorted = sort(distance,"descend");
for i=1:19
    index = find(distance==sorted(i));
    [m,n] = size(index);
    if m == 3
        fprintf("Song number %d, %d and %d are %f close to song 1\n",index,sorted(i))
    elseif m == 2
        fprintf("Song number %d and %d are %f close to song 1\n",index,sorted(i))
    else
        fprintf("Song number %d is %f close to song 1\n",index,sorted(i))
    end
end

Answer 1

[Matt J]

Crescendo = [0 1 0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 0]';
Descrescendo = [1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 0 0]';
Staccato = [0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0]';
Portamento = [0 0 0 1 0 1 1 1 0 0 0 1 1 0 0 1 0 1 0 0]';
Tempo = [1 1 1 0 1 0 1 0 0 1 1 1 1 0 1 0 0 0 1 1]';
Intervals = [0 1 1 1 1 1 0 1 1 0 0 0 1 0 0 1 1 1 1 1]';
M = [Crescendo Descrescendo Staccato Portamento Tempo Intervals];
S = M*M';
[U,W,V] = svds(S,2);

U=normalize(U,2,'n');

distance = U(2:19,:)*U(1,:).';

sorted=flip(unique(distance));

for i=1:numel(sorted)
    index = find(distance==sorted(i));
    m=numel(index);
    if m == 3
        fprintf("Song number %d, %d and %d are %f close to song 1\n",index,sorted(i))
    elseif m == 2
        fprintf("Song number %d and %d are %f close to song 1\n",index,sorted(i))
    else
        fprintf("Song number %d is %f close to song 1\n",index,sorted(i))
    end
end

Song number 18 is 0.999745 close to song 1
Song number 14 is 0.997055 close to song 1

Song number 2 and 4 are 0.971605 close to song 1

Song number 9 is 0.967074 close to song 1
Song number 10 is 0.948000 close to song 1
Song number 1 is 0.763357 close to song 1
Song number 12 is 0.695057 close to song 1
Song number 13 is 0.527134 close to song 1

Song number 6 and 11 are 0.424546 close to song 1

Song number 8 is 0.302743 close to song 1
Song number 16 is 0.153338 close to song 1
Song number 5 is -0.280355 close to song 1

Song number 3, 7 and 17 are -0.345152 close to song 1

Song number 15 is -0.503935 close to song 1