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view main/statistics/inst/normalise_distribution.m @ 9701:b4dccb6ed82d octave-forge
normalise_distribution: workaround due to octave bug #34765 was made for releases 3.4.X only. However, regression in octave is not yet fixed as of 3.6.1. Expanding workaround for all releases after 3.4.0 until fixed in core.
author | carandraug |
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date | Wed, 14 Mar 2012 22:29:00 +0000 |
parents | c115cdd2c118 |
children | 3828294802c5 |
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## Copyright (C) 2011 Alexander Klein <alexander.klein@math.uni-giessen.de> ## ## 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{Function File} {@var{NORMALISED} =} normalise_distribution (@var{DATA}) ## @deftypefnx{Function File} {@var{NORMALISED} =} normalise_distribution (@var{DATA}, @var{DISTRIBUTION}) ## @deftypefnx{Function File} {@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 ( "First argument must be a vector or matrix" ); end 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 ) distribution = { str2func( distribution ) }; else error ( ["Second argument cannot contain more than one string" ... " unless in a cell array"] ); end ##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 ); end ##Does it eventually consist of function handles only if ( !all ( cellfun ( @ ( h ) ( strcmp ( typeinfo ( h ), typeinfo ( @(x)(x) ) ) ), distribution ) ) ) error ( ["Second argument must contain either" ... " a single function name or handle or " ... " a cell array of either all function names or handles!"] ); end else error ( "Illegal second argument: ", typeinfo ( distribution ) ); end end else distribution = []; end if ( nargin == 3 ) if ( !isscalar ( dimension ) || ( dimension != 1 && dimension != 2 ) ) error ( "Third argument must be either 1 or 2" ); end else if ( isvector ( data ) && rows ( data ) == 1 ) dimension = 2; else dimension = 1; end end trp = ( dimension == 2 ); if ( trp ) data = data'; end 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) ); end normal = precomputed_normalisation; end #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. ## A regression in the 3.4 series made this no longer work so we behave ## differently depending on octave version. This applies the fix for all ## 3.4 releases but it probably only appeared on 3.4.3 (can someone check?) ## See https://savannah.gnu.org/bugs/index.php?34765 ## Turns out that the bug was not completely fixed the first time and is ## still present in 3.6.1 ## FIXME Once package dependency increases beyond an octave version that ## has this fixed, remove this if (compare_versions (OCTAVE_VERSION, "3.4", "<")) ## this is how it should work f_remap = @( k ) ( normal ( k ) ); normalised ( :, k ) = arrayfun ( f_remap, target_indices ); else ## this is the workaround because of bug in 3.4.?? for index = 1:numel(target_indices) normalised ( index, k ) = normal(target_indices(index)); endfor endif end 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 ) ); end else error ( "Number of distributions does not match data size! ") end end if ( trp ) normalised = normalised'; end endfunction %!test %! v = normalise_distribution ( [ 1 2 3 ], [], 1 ); %! assert ( v, [ 0 0 0 ] ) %!test %! v = normalise_distribution ( [ 1 2 3 ], [], 2 ); %! assert ( v, norminv ( [ 1 3 5 ] / 6 ), 3 * eps ) %!test %! v = normalise_distribution ( [ 1 2 3 ]', [], 2 ); %! assert ( v, [ 0 0 0 ]' ) %!test %! v = normalise_distribution ( [ 1 2 3 ]' , [], 1 ); %! assert ( v, norminv ( [ 1 3 5 ]' / 6 ), 3 * eps ) %!test %! v = normalise_distribution ( [ 1 1 2 2 3 3 ], [], 2 ); %! assert ( v, norminv ( [ 3 3 7 7 11 11 ] / 12 ), 3 * eps ) %!test %! v = normalise_distribution ( [ 1 1 2 2 3 3 ]', [], 1 ); %! assert ( v, norminv ( [ 3 3 7 7 11 11 ]' / 12 ), 3 * eps ) %!test %! A = randn ( 10 ); %! N = normalise_distribution ( A, @normcdf ); %! assert ( A, N, 1000 * eps ) %!xtest %! A = exprnd ( 1, 100 ); %! N = normalise_distribution ( A, @ ( x ) ( expcdf ( x, 1 ) ) ); %! assert ( mean ( vec ( N ) ), 0, 0.1 ) %! assert ( std ( vec ( N ) ), 1, 0.1 ) %!xtest %! A = rand (1000,1); %! N = normalise_distribution ( A, "unifcdf" ); %! assert ( mean ( vec ( N ) ), 0, 0.1 ) %! assert ( std ( vec ( N ) ), 1, 0.1 ) %!xtest %! A = [rand(1000,1), randn( 1000, 1)]; %! N = normalise_distribution ( A, { "unifcdf", "normcdf" } ); %! assert ( mean ( N ), [ 0, 0 ], 0.1 ) %! assert ( std ( N ), [ 1, 1 ], 0.1 ) %!xtest %! A = [rand(1000,1), randn( 1000, 1), exprnd( 1, 1000, 1 )]'; %! N = normalise_distribution ( A, { @unifcdf; @normcdf; @( x )( expcdf ( x, 1 ) ) }, 2 ); %! assert ( mean ( N, 2 ), [ 0, 0, 0 ]', 0.1 ) %! assert ( 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 ( mean ( N ), [ 0 0 0 0.38 0.38 0.38 0 0 0 ], 0.1 ); %! assert ( var ( N ), [ 1 1 1 2.59 2.59 2.59 1 1 1 ], 0.1 ); %!test %! fail ("normalise_distribution( zeros ( 3, 4 ), { @unifcdf; @normcdf; @( x )( expcdf ( x, 1 ) ) } )", ... %! "Number of distributions does not match data size!");