%REORDERCLASSES Reorder the lablist
%
% X = REORDERCLASSES(X,LABLIST)
% X = REORDERCLASSES(X,I)
%
% INPUT
% X (labeled) dataset
% LABLIST correctly ordered lablist
% I permutation vector
%
% OUTPUT
% X dataset with reordered classes
%
% DESCRIPTION
% Change the order of the classes in dataset X to the one defined in
% LABLIST. When only a single class label is given in LABLIST, this
% class is put in the first position, and the others are just kept in
% the order they already were.
%
% For example, when I have a dataset A with a labellist {'apple'
% 'banana' 'pear'} and I perform B=REORDERCLASS(A,3) (or identically
% B=REORDERCLASS(A,[3 1 2])), I will get a dataset B with a new
% labellist of {'pear' 'apple' 'banana'}.
%
% SEE ALSO (<a href="http://37steps.com/prtools">PRTools Guide</a>)
% RENUMLAB
% Copyright: D.M.J. Tax, D.M.J.Tax@37steps.com
% Faculty EWI, Delft University of Technology
% P.O. Box 5031, 2600 GA Delft, The Netherlands
function x = reorderclasses(x,lablist)
% First deal with the situation that lablist is a cellarray:
if isa(lablist,'cell')
if isa(lablist{1},'char')
lablist = strvcat(lablist);
else % it should be a double...
if ~isa(lablist,'double')
error('Only stringlabels or numeric labels are allowed.');
end
lablist = cell2mat(lablist);
end
end
% ... and make sure numerical labels are column vectors
if isa(lablist,'double')
lablist = lablist(:);
end
ll = getlablist(x);
c = size(ll,1);
% Second deal with the situation that we only supplied a single 'target
% class':
if size(lablist,1)==1
if isa(lablist,'char') % the single label is a string:
matchnr = strmatch(lablist,ll);
if isempty(matchnr)
error('Cannot find %s in the dataset.',lablist);
end
else % the single label is a number:
% is this number a class label, or the index in the lablist??
if (lablist<1) | (lablist>c)
% probably a class label
matchnr = find(lablist==ll);
if isempty(matchnr)
error('Cannot find %d in the dataset.',lablist);
end
else
matchnr = lablist;
end
end
% make a complete permutation matrix:
lablist = (1:c)';
lablist(matchnr) = [];
lablist = [matchnr; lablist];
end
% test to see if lablist has at least enough elements:
if size(lablist,1)~=c
error('New lablist or permutation vector does not match with number of classes.');
end
% Do the matching and reordering, depending if the class labels in X are
% string or numerical
if isa(ll,'char') % X has string labels
if isa(lablist,'char') % we have to match 2 string labels
I = zeros(c,1);
for i=1:c
matchnr = strmatch(lablist(i,:),ll);
if isempty(matchnr)
error('Label %s cannot be found in the dataset.',lablist(i,:));
end
I(i) = matchnr;
end
else % we have to apply a simple permutation on strings in X
if ~isa(lablist,'double')
error('Expecting a permutation label or a new lablist.');
end
I = lablist;
if (min(I)<1) | (max(I)>c)
error('Not a valid permutation vector is supplied.');
end
lablist = ll(I,:);
end
else % dataset X has numeric labels
if (min(lablist)<1) | (max(lablist)>c)
% we are probably NOT dealing with a permutation vector
I = zeros(c,1);
for i=1:c
matchnr = find(lablist(i)==ll);
if isempty(matchnr)
error('Label %d cannot be found in the dataset.',lablist(i));
end
I(i) = matchnr;
end
else % just a permutation on the numeric labels
I = lablist;
lablist = ll(I,:);
end
end
% So finally perform the transformation:
x = setlablist(x, lablist);
[sortedI,I] = sort(I); %kind of inverse the I
% and be careful with unlabeled data
nlab = getnlab(x);
J = find(nlab>0);
nlab(J) = I(nlab(J));
x = setnlab(x, nlab);
return