How to sort a matrix with the second row depending on the sorting of the first row?

Hey guys,

I have 2 vectors, time and signal, with corresponding vectors [ 1 3 2 4 7] and [12 14 11 13 16]. I have this quicksort code:

function dataOut = Quicksort(data)
lenD = size(data,2);
ind = cast(floor(lenD/2),'uint8');
j = 1;
k = 1;
L = [];
R = [];
if(lenD<2)
  dataOut = data;
else
  pivot = data(ind);
  for i=1:lenD
    if(i~=ind)
      if(data(i)<pivot)
        L(j) = data(i);
        j = j+1;
      else
        R(k) = data(i);
        k = k+1;
      end
    end
  end
  L = Quicksort(L);
  R = Quicksort(R);
  dataOut = [L pivot R];
end

---------------------------------

It works well for only 1 vector, but what I ideally want is the 2nd vector to move in correspondance with the first, i.e the sorted vectors become: [1 2 3 4 7] and [12 11 14 13 16]. I cannot use the sort command. How do I go about this???

Kind regards,

Tom

Answer 1

[Stephen23]

One easy way to obtain indices from any sorting algorithm is to "hang" some indices onto the data before sorting, and apply exactly the same permutations to those indices as are applied to the data themselves. I wrapped your function in another one to make this a bit easier to implement:

function [out,idx] = quicksort(vec)
assert(isrow(vec),'Input must be a row vector')
Z = qsRecFun([vec;1:numel(vec)]);
out = Z(1,:);
idx = Z(2,:);
end
function Sm = qsRecFun(M)
N = size(M,2);  
if N<2
    Sm = M;
else
    jj = 1;
    kk = 1;
    Lm = [];
    Rm = [];
    X = fix(N/2);
    P = M(:,X);
    for ii = [1:X-1,X+1:N]
        if M(1,ii)<P(1)
            Lm(:,jj) = M(:,ii);
            jj = jj+1;
        else
            Rm(:,kk) = M(:,ii);
            kk = kk+1;
        end
    end
    Lm = qsRecFun(Lm);
    Rm = qsRecFun(Rm);
    Sm = [Lm,P,Rm];
end
end

And checking the output against that of MATLAB's sort:

>> V = randi(9,1,7)
V =
   9   5   3   8   1   6   2
>> [Y,X] = quicksort(V)
Y =
   1   2   3   5   6   8   9
X =
   5   7   3   2   6   4   1
>> [Y,X] = sort(V)
Y =
   1   2   3   5   6   8   9
X =
   5   7   3   2   6   4   1
   

Note that you could also improve the code by using logical indexing, although this would look less like the naive algorithm itself.

[Thomas Nell]

Stephen, you're a genius, thanks so much man.