Calculate differences between all values in vector

I'm trying to produce some code that will calculate the differences between nearly all values in a vector.

Specifically, say I have a vector [2 3 1 4]

Starting at 2 and moving through the vector, I need to calculate the difference between 2 and 3 (i.e. -1), 2 and 1, 2 and 4, then 3 and 1, 3 and 4, and then 1 and 4. I don't need to calculate the difference between, say, 3 and 2, and I'm only interested in the differences in one direction.

The vector I want to use the code on will be substantially bigger, and I'd like the output stored in a new, single column vector. I figured diff might provide a way forward, but can't see how to implement it specifying a 'lag'. Any help would be appreciated.

Answer 1

[David Sanchez]

my_vct =  [2 3 1 4 5 6];
L = numel(my_vct);
tmp_vct = [my_vct(2:end) my_vct(end)];

x = my_vct - tmp_vct;

sol = zeros(L-1,L);
for k=2:L
    tmp_vct = [my_vct(k:end) my_vct((L-k+2):end)];
    sol(k-1,:) = my_vct - tmp_vct;

end

sol =

      -1     2    -3    -1    -1     0
       1    -1    -4    -2     0     0
      -2    -2    -5     0     0     0
      -3    -3     0     0     0     0
      -4     0     0     0     0     0

[David Sanchez]

First definition of tmp_vct and x are not needed:

%%%%%%%%%%%%%%%%%%%%
my_vct =  [2 3 1 4 5 6];
L = numel(my_vct);

sol = zeros(L-1,L);
for k=2:L
    tmp_vct = [my_vct(k:end) my_vct((L-k+2):end)];
    sol(k-1,:) = my_vct - tmp_vct;

    end
  %%%%%%%%%%%%%%%%%%%%

sol =

        -1     2    -3    -1    -1     0
         1    -1    -4    -2     0     0
        -2    -2    -5     0     0     0
        -3    -3     0     0     0     0
        -4     0     0     0     0     0

[Swisslog]

...and

sol=sol(:);
sol(sol==0) = [];

to reduce to a single column I guess. Many thanks!

[Jos (10584)]

This will be very slow for large inputs.

Answer 2

[Jos (10584)]

V = [2 3 1 4] ;

D1 = bsxfun(@minus,V(:), V(:).') % square form

% another option, only unique combinations, requires less memory
D2 = arrayfun(@(k) V(k:end)-V(k), 1:numel(k),'un',0)
D2 = [D2{:}]

[Ondrej]

There is a small bug I believe. It should be

D2 = arrayfun(@(k) V(k:end)-V(k), 1:numel(V),'un',0)

[Soyy Tuffjefe]

Thanks to Jos (10584) for your code...

I have a 5x1000 matrix (numerical entries),

M=[1 5 8 75 120;

1 25 18 5 10;

⋮ ⋮ ⋮ ⋮ ⋮

7 39 118 125 10]

and want to apply her or his code:

D2 = arrayfun(@(k) V(k:end)-V(k), 1:numel(k),'un',0)

D2 = [D2{:}

to every row of M, please, could anybody modify this code for this job.

Thanks in avanced!

[Mehri Mehrnia]

A great idea.

any body knows computation cost of bsxfun(@minus,V(:), V(:).')???

I mean linear or 2nd order or...

I work with arrays in order of millions, that's why it's important for me.

Answer 3

[Mischa Kim]

Swisslog, you could use

 pdist([2 3 1 4]',@(x,y) x-y)

[Swisslog]

Sounds ideal! Afraid I don't have the statistics toolbox though