## Copyright (C) 2011 Alexander Klein <alexander.klein@math.uni-giessen.de>
## Copyright (C) 2022-2023 Andreas Bertsatos <abertsatos@biol.uoa.gr>
##
## This file is part of the statistics package for GNU Octave.
##
## This program is free software; you can redistribute it and/or modify it under
## the terms of the GNU General Public License as published by the Free Software
## Foundation; either version 3 of the License, or (at your option) any later
## version.
##
## This program is distributed in the hope that it will be useful, but WITHOUT
## ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
## FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more
## details.
##
## You should have received a copy of the GNU General Public License along with
## this program; if not, see <http://www.gnu.org/licenses/>.

## -*- texinfo -*-
## @deftypefn  {statistics} {@var{NORMALISED} =} normalise_distribution (@var{DATA})
## @deftypefnx {statistics} {@var{NORMALISED} =} normalise_distribution (@var{DATA}, @var{DISTRIBUTION})
## @deftypefnx {statistics} {@var{NORMALISED} =} normalise_distribution (@var{DATA}, @var{DISTRIBUTION}, @var{DIMENSION})
##
## Transform a set of data so as to be N(0,1) distributed according to an idea
## by van Albada and Robinson.
##
## This is achieved by first passing it through its own cumulative distribution
## function (CDF) in order to get a uniform distribution, and then mapping
## the uniform to a normal distribution.
##
## The data must be passed as a vector or matrix in @var{DATA}.
## If the CDF is unknown, then [] can be passed in @var{DISTRIBUTION}, and in
## this case the empirical CDF will be used.
## Otherwise, if the CDFs for all data are known, they can be passed in
## @var{DISTRIBUTION},
## either in the form of a single function name as a string,
## or a single function handle,
## or a cell array consisting of either all function names as strings,
## or all function handles.
## In the latter case, the number of CDFs passed must match the number
## of rows, or columns respectively, to normalise.
## If the data are passed as a matrix, then the transformation will
## operate either along the first non-singleton dimension,
## or along @var{DIMENSION} if present.
##
## Notes:
## The empirical CDF will map any two sets of data
## having the same size and their ties in the same places after sorting
## to some permutation of the same normalised data:
## @example
## @code{normalise_distribution([1 2 2 3 4])}
## @result{} -1.28  0.00  0.00  0.52  1.28
##
## @code{normalise_distribution([1 10 100 10 1000])}
## @result{} -1.28  0.00  0.52  0.00  1.28
## @end example
##
## Original source:
## S.J. van Albada, P.A. Robinson
## "Transformation of arbitrary distributions to the
## normal distribution with application to EEG
## test-retest reliability"
## Journal of Neuroscience Methods, Volume 161, Issue 2,
## 15 April 2007, Pages 205-211
## ISSN 0165-0270, 10.1016/j.jneumeth.2006.11.004.
## (http://www.sciencedirect.com/science/article/pii/S0165027006005668)
## @end deftypefn

function normalised = normalise_distribution (data, distribution, dimension)

  if (nargin < 1 || nargin > 3)
    print_usage;
  elseif (! ismatrix (data) || length (size (data)) > 2)
    error (strcat ("normalise_distribution: first argument", ...
                   " must be a vector or matrix."));
  endif

  if (nargin >= 2)
    if (! isempty (distribution))
      ## Wrap a single handle in a cell array.
      if (strcmp (typeinfo (distribution), typeinfo (@(x)(x))))

        distribution = {distribution};

      ## Do we have a string argument instead?
      elseif (ischar (distribution))

        ## Is it a single string?
        if (rows (distribution) == 1)
          temp = str2func ([distribution]);
          distribution = {temp};
        else
          error (strcat ("normalise_distribution: second argument cannot", ...
                         " contain more than one string unless in a cell", ...
                         " array."));
        endif

      ## Do we have a cell array of distributions instead?
      elseif (iscell (distribution))

        ## Does it consist of strings only?
        if (all (cellfun (@ischar, distribution)))
            distribution = cellfun (@str2func, distribution, ...
                                    'UniformOutput', false );
        endif

        ## Does it eventually consist of function handles only
        if (! all (cellfun (@(h) (strcmp (typeinfo (h), typeinfo ...
                           (@(x)(x)))), distribution)))

          error (strcat ("normalise_distribution: second argument must", ...
                         " contain either a single function name or", ...
                         " handle or a cell array of either all function", ...
                         " names or handles!"));
        endif
      else
        error ( "Illegal second argument: ", typeinfo ( distribution ) );
      endif
    endif
  else
    distribution = [];
  endif

  if (nargin == 3)
    if (! isscalar (dimension) || (dimension != 1 && dimension != 2))
      error ("normalise_distribution: third argument must be either 1 or 2.");
    endif
  else
    if (isvector (data) && rows (data) == 1)
      dimension = 2;
    else
      dimension = 1;
    endif
  endif

  trp = (dimension == 2);

  if (trp)
    data = data';
  endif

  r = rows (data);
  c = columns (data);
  normalised = NA (r, c);

  ## Do we know the distribution of the sample?
  if (isempty (distribution))
    precomputed_normalisation = [];
    for k = 1 : columns ( data )
      ## Note that this line is in accordance with equation (16) in the
      ## original text. The author's original program, however, produces
      ## different values in the presence of ties, namely those you'd
      ## get replacing "last" by "first".
      [uniq, indices] = unique (sort (data(:, k)), 'last');

      ## Does the sample have ties?
      if (rows (uniq) != r)
        ## Transform to uniform, then normal distribution.
        uniform = ( indices - 1/2 ) / r;
        normal = norminv ( uniform );
      else
        ## Without ties everything is pretty much straightforward as
        ## stated in the text.
        if (isempty (precomputed_normalisation))
          precomputed_normalisation = norminv (1 / (2*r) : 1/r : 1 - 1 / (2*r));
        endif
        normal = precomputed_normalisation;
      endif
      ## Find the original indices in the unsorted sample.
      ## This somewhat quirky way of doing it is still faster than
      ## using a for-loop.
      [ ignore, ignore, target_indices ] = unique ( data(:, k ) );

      ## Put normalised values in the places where they belong.
      normalised( :, k ) = normal(target_indices);
    endfor
  else
    ## With known distributions, everything boils down to a few lines of code

    ##The same distribution for all data?
    if (all (size (distribution) == 1))
      normalised = norminv (distribution{1,1}(data));
    elseif (length (vec (distribution)) == c)
      for k = 1 : c
        normalised(:, k) = norminv (distribution{k}(data)(:, k));
      endfor
    else
      error (strcat ("normalise_distribution: number of distributions", ...
                     " does not match data size!"));
    endif
  endif

  if (trp)
    normalised = normalised';
  endif
endfunction

%!test
%! v = normalise_distribution ([1 2 3], [], 1);
%! assert_equal (v, [0 0 0])

%!test
%! v = normalise_distribution ([1 2 3], [], 2);
%! assert_equal (v, norminv ([1 3 5] / 6), 3 * eps)

%!test
%! v = normalise_distribution ([1 2 3]', [], 2);
%! assert_equal (v, [0 0 0]')

%!test
%! v = normalise_distribution ([1 2 3]', [], 1);
%! assert_equal (v, norminv ([1 3 5]' / 6), 3 * eps)

%!test
%! v = normalise_distribution ([1 1 2 2 3 3], [], 2);
%! assert_equal (v, norminv ([3 3 7 7 11 11] / 12), 3 * eps)

%!test
%! v = normalise_distribution ([1 1 2 2 3 3]', [], 1);
%! assert_equal (v, norminv ([3 3 7 7 11 11]' / 12), 3 * eps)

%!test
%! A = randn ( 10 );
%! N = normalise_distribution (A, @normcdf);
%! assert_equal (A, N, 10000 * eps)

%!test
%! A = exprnd (1, 100);
%! N = normalise_distribution (A, @(x)(expcdf (x, 1)));
%! assert_equal (mean (vec (N)), 0, 0.1)
%! assert_equal (std (vec (N)), 1, 0.1)

%!test
%! A = rand (1000,1);
%! N = normalise_distribution (A, {@(x)(unifcdf (x, 0, 1))});
%! assert_equal (mean (vec (N)), 0, 0.2)
%! assert_equal (std (vec (N)), 1, 0.1)

%!test
%! A = [rand(1000,1), randn(1000, 1)];
%! N = normalise_distribution (A, {@(x)(unifcdf (x, 0, 1)), @normcdf});
%! assert_equal (mean (N), [0, 0], 0.2)
%! assert_equal (std (N), [1, 1], 0.1)

%!test
%! A = [rand(1000,1), randn(1000, 1), exprnd(1, 1000, 1)]';
%! N = normalise_distribution (A, {@(x)(unifcdf (x, 0, 1)); @normcdf; @(x)(expcdf (x, 1))}, 2);
%! assert_equal (mean (N, 2), [0, 0, 0]', 0.2);
%! assert_equal (std (N, [], 2), [1, 1, 1]', 0.1);

%!xtest
%! A = exprnd (1, 1000, 9); A(300:500, 4:6) = 17;
%! N = normalise_distribution (A);
%! assert_equal (mean (N), [0 0 0 0.38 0.38 0.38 0 0 0], 0.1);
%! assert_equal (var (N), [1 1 1 2.59 2.59 2.59 1 1 1], 0.1);

%!test
%!error normalise_distribution (zeros (3, 4), ...
%! {@(x)(unifcdf (x, 0, 1)); @normcdf; @(x)(expcdf (x,1))});
