Is it possible to sort an array by making the repeated elements in the last?

Greetings.

Consider a vector A = [-2 -2 -1 3 3], is it possible to sort this vector (a generalized method) to [-1 -2 -2 -3 -3]?

The only thing that matters is that the repeated elements are in the last, it doesn't matter what way the repeated elements themselves are sorted (for example, [-1 -2 -2 -3 -3] or [-1 -3 -3 -2 -2] are fine) I just want the distinct value to be on the left before any repeated value.

Thank you in advance,

M. Ayoub

Answer 1

[Stephen23]

>> A = [-2,-2,-1,3,3,3,9,9,0,99]
A =
   -2   -2   -1    3    3    3    9    9    0   99
>> [cnt,idx] = histc(A,unique(A));
>> [~,idy] = sort(cnt>1);
>> [~,idz] = ismember(idx,idy);
>> [~,ids] = sort(idz);
>> B = A(ids)
B =
   -1    0   99   -2   -2    3    3    3    9    9

And

>> A = [-1,-1,-3,-5,-5]
...
B =
  -3  -5  -5  -1  -1

[Jan]

This sorts in unique, twice in sort and again in ismember. In addition the perfectly working histc is deprecated (what a pity!).

[Mohammad Ayoub]

Thank you Stephen and Jan.

Jan I know your answer is true but I think it is a little advanced for me haha, my code is really small and I am still new to MATLAB, so what Stephen wrote works perfectly fine in my application!

By the way, can anyone explain to me what happened exactly (step by step or write the code again with comments) in the code? If you please, so I can understand it better.

Thank you very much!

Answer 2

[KSSV]

A = [-2 -2 -1 3 3] ;
s = sign(A) ;
[val,idx] = sort(abs(A)) ;
iwant = s(idx).*val

[Mohammad Ayoub]

Thank you for your time. Your answer completely works on my example, however I want a generalized form :( I need it to work on any array that contains repeated elements. Is it possible to do that?

*Note: I tried [-1 -1 -3 -5 -5] with your solution but it didn't work.

Thank you!

[KSSV]

What result you expect for [-1 -1 -3 -5 -5]?

[Mohammad Ayoub]

The result I expect is [-3 -1 -1 -5 -5] or [-3 -5 -5 -1 -1]

The important thing is that the -3 (which is not repeated) is in the beginning of the array.

By the way, I am limiting my code to 1 repeated value, there will be no more than one repeated element in the arrays (I think this makes it slightly easier)

Answer 3

[Jan]

With FEX: RunLength:

A      = [-1 -1 -3 -5 -5];
[b, n] = RunLength(sort(A));
m      = (n == 1);
result = [b(m), RunLength(b(~m), n(~m)];

If you do not have a C-compiler installed, use RunLength_M from this submission.

Maybe this is faster:

sA            = sort(A);
[b, n, index] = RunLength(sA);
mask          = false(size(A));
mask(index(n == 1)) = true;
result        = [sA(mask), sA(~mask)];