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## -*- texinfo -*-
## @deftypefn  {} {@var{tau} =} kendall (@var{x})
## @deftypefnx {} {@var{tau} =} kendall (@var{x}, @var{y})
## @cindex Kendall's Tau
## Compute Kendall's
## @tex
## $\tau$.
## @end tex
## @ifnottex
## @var{tau}.
## @end ifnottex
##
## For two data vectors @var{x}, @var{y} of common length @math{N}, Kendall's
## @tex
## $\tau$
## @end tex
## @ifnottex
## @var{tau}
## @end ifnottex
## is the correlation of the signs of all rank differences of
## @var{x} and @var{y}; i.e., if both @var{x} and @var{y} have distinct
## entries, then
## @tex
## $$ \tau = {1 \over N(N-1)} \sum_{i,j} {\rm sign}(q_i-q_j) \, {\rm sign}(r_i-r_j) $$
## @end tex
## @ifnottex
##
## @example
## @group
##          1
## @var{tau} = -------   SUM sign (@var{q}(i) - @var{q}(j)) * sign (@var{r}(i) - @var{r}(j))
##       N (N-1)   i,j
## @end group
## @end example
##
## @end ifnottex
## @noindent
## in which the
## @tex
## $q_i$ and $r_i$
## @end tex
## @ifnottex
## @var{q}(i) and @var{r}(i)
## @end ifnottex
## are the ranks of @var{x} and @var{y}, respectively.
##
## If @var{x} and @var{y} are drawn from independent distributions,
## Kendall's
## @tex
## $\tau$
## @end tex
## @ifnottex
## @var{tau}
## @end ifnottex
## is asymptotically normal with mean 0 and variance
## @tex
## ${2 (2N+5) \over 9N(N-1)}$.
## @end tex
## @ifnottex
## @code{(2 * (2N+5)) / (9 * N * (N-1))}.
## @end ifnottex
##
## @code{kendall (@var{x})} is equivalent to @code{kendall (@var{x},
## @var{x})}.
## @seealso{ranks, spearman}
## @end deftypefn

function tau = kendall (x, y = [])

  if (nargin < 1)
    print_usage ();
  endif

  if (   ! (isnumeric (x) || islogical (x))
      || ! (isnumeric (y) || islogical (y)))
    error ("kendall: X and Y must be numeric matrices or vectors");
  endif

  if (ndims (x) != 2 || ndims (y) != 2)
    error ("kendall: X and Y must be 2-D matrices or vectors");
  endif

  if (isrow (x))
    x = x.';
  endif
  [n, c] = size (x);

  if (nargin == 2)
    if (isrow (y))
      y = y.';
    endif
    if (rows (y) != n)
      error ("kendall: X and Y must have the same number of observations");
    else
      x = [x, y];
    endif
  endif

  if (isa (x, "single") || isa (y, "single"))
    cls = "single";
  else
    cls = "double";
  endif
  r   = ranks (x);
  m   = sign (kron (r, ones (n, 1, cls)) - kron (ones (n, 1, cls), r));
  tau = corr (m);

  if (nargin == 2)
    tau = tau(1 : c, (c + 1) : columns (x));
  endif

endfunction


%!test
%! x = [1:2:10];
%! y = [100:10:149];
%! assert (kendall (x,y), 1, 5*eps);
%! assert (kendall (x,fliplr (y)), -1, 5*eps);

%!assert (kendall (1), NaN)
%!assert (kendall (single (1)), single (NaN))

## Test input validation
%!error <Invalid call> kendall ()
%!error kendall (['A'; 'B'])
%!error kendall (ones (2,1), ['A'; 'B'])
%!error kendall (ones (2,2,2))
%!error kendall (ones (2,2), ones (2,2,2))
%!error kendall (ones (2,2), ones (3,2))
