Apply function to all rows

Question

I have a function, ranker, that takes a vector and assigns numerical ranks to it in ascending order. For example,
ranker([5 1 3 600]) = [3 1 2 4] or
ranker([42 300 42 42 1 42] = [3.5 6 3.5 3.5 1 3.5] .

I am using a matrix, variable_data and I want to apply the ranker function to each row for all rows in variable data. This is my current solution, but I feel there is a way to vectorize it and have it as equally fast :p

variable_ranks = nan(size(variable_data));
for i=1:1:numel(nmac_ids)
    variable_ranks(i,:) = ranker(abs(variable_data(i,:)));
end

Answer 1

With collaboration from Amro and Jonas

variable_ranks = tiedrank(variable_data')';

Ranker has been replaced by the Matlab function in the Stat toolbox (sorry for those who don't have it),

[R,TIEADJ] = tiedrank(X) computes the ranks of the values in the vector X. If any X values are tied, tiedrank computes their average rank. The return value TIEADJ is an adjustment for ties required by the nonparametric tests signrank and ranksum, and for the computation of Spearman's rank correlation.

TIEDRANK will compute along columns in Matlab 7.9.0 (R2009b), however it is undocumented. So by transposing the input matrix, rows turn into columns and will rank them. The second transpose is then used to organize the data in the same manner as the input. There in essence is a very classy hack :p

Answer 2

If you place the matrix rows into a cell array, you can then apply a function to each cell.

Consider this simple example of applying the SORT function to each row

a = rand(10,3);
b = cell2mat( cellfun(@sort, num2cell(a,2), 'UniformOutput',false) );
%# same as: b = sort(a,2);

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You can even do this:

b = cell2mat( arrayfun(@(i) sort(a(i,:)), 1:size(a,1), 'UniformOutput',false)' );

Again, you version with the for loop is probably faster..

Answer 3

One way would be to rewrite ranker to take array input

sizeData = size(variable_data);

[sortedData,almostRanks] = sort(abs(variable_data),2);
[rowIdx,colIdx] = ndgrid(1:sizeData(1),1:sizeData(2));
linIdx = sub2ind(sizeData,rowIdx,almostRanks);
variable_ranks = variable_data;
variable_ranks(linIdx) = colIdx;

%# break ties by finding subsequent equal entries in sorted data
[rr,cc] = find(diff(sortedData,1,2) == 0);
ii = sub2ind(sizeData,rr,cc);
ii2 = sub2ind(sizeData,rr,cc+1);
ii = sub2ind(sizeData,rr,almostRanks(ii));
ii2 = sub2ind(sizeData,rr,almostRanks(ii2));
variable_ranks(ii) = variable_ranks(ii2);

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EDIT

Instead, you can just use TIEDRANK from TMW (thanks, @Amro):

variable_rank = tiedrank(variable_data')';

Answer 4

I wrote a function that does this, it's on the FileExchange tiedrank_(X,dim). And it looks like this...

%[Step 0a]: force dim to be 1, and compress everything else into a single 
%dimension. We will reverse this process at the end.
if dim > 1 
    otherDims = 1:length(size(X));
    otherDims(dim) = [];
    perm = [dim otherDims];
    X = permute(X,perm);
end
originalSiz = size(X);
X = reshape(X,originalSiz(1),[]);
siz = size(X);

%[Step 1]: sort and get sorting indicies
[X,Ind] = sort(X,1);

%[Step 2]: create matrix [D], which has +1 at the start of consecutive runs
% and -1 at the end, with zeros elsewhere.
D = zeros(siz,'int8');
D(2:end-1,:) = diff(X(1:end-1,:) == X(2:end,:));
D(1,:) = X(1,:) == X(2,:);
D(end,:) = -( X(end,:) == X(end-1,:) );

clear X

%[Step 3]: calculate the averaged rank for each consecutive run
[a,~] = find(D);
a = reshape(a,2,[]);
h = sum(a,1)/2;

%[Step 4]: insert the troublseome ranks in the relevant places
L = zeros(siz);
L(D==1) = h;
L(D==-1) = -h;
L = cumsum(L);
L(D==-1) = h; %cumsum set these ranks to zero, but we wanted them to be h

clear D h

%[Step 5]: insert the simple ranks (i.e. the ones that didn't clash)
[L(~L),~] = find(~L);

%[Step 6]: assign the ranks to the relevant position in the matrix
Ind = bsxfun(@plus,Ind,(0:siz(2)-1)*siz(1)); %equivalent to using sub2ind + repmat
r(Ind) = L;

%[Step 0b]: As promissed, we reinstate the correct dimensional shape and order
r = reshape(r,originalSiz);
if dim > 1
    r = ipermute(r,perm);
end

I hope that helps someone.