ranking (ordering) values with repeats

Hello Community,

Im hoping some of you have a clever solution to this problem. Im looking for fast and efficient way to rank (order)a vector of numbers in a particular way when repeated values arise.

To make it simple, suppose I have a row vector:

data = [-1 2 0 -2 0]

I know I can rank them using the 3rd output of "unique":

>> [~,~,rnk] = unique(data)

rnk =

       2     4     3     1     3

What I like about this is that it assigns the same rank to the repeated zeros. What I don't like about this is that the top rank is now "4" even though I have 5 values. I would prefer this:

>> rnk = myrank(data)

rnk =

       2     5     3     1     3

Ive also played around with the second output of "sort" quite a bit, but since this output produces indicies of the sorted values within the original array, there is no simple way (that I've found) to associate the same rank with repeated values.

Im just wondering if there is something simple that Im missing.

Thanks!

Answer 1

[Oleg Komarov]

If you have the Statistics Toolbox, but it's kinda an overkill:

floor(tiedrank([-1 2 0 -2 0]))
ans =
     2     5     3     1     3

Otherwise:

data   = [-1 2 0 -2 0];

% Sort data
[srt, idxSrt]  = sort(data);
% Find where are the repetitions
idxRepeat      = [false diff(srt) == 0];
% Rank with tieds but w/o skipping
rnkNoSkip      = cumsum(~idxRepeat);
% Preallocate rank
rnk            = 1:numel(data);
% Adjust for tieds (and skip)
rnk(idxRepeat) = rnkNoSkip(idxRepeat);
% Sort back
rnk(idxSrt)    = rnk

rnk =
   2     5     3     1     3

[owr]

This is very nice Oleg, thank you for taking the time. I do have the Statistics Toolbox and had forgotten about "tiedrank". I tried that when I first worked on this problem about a year ago, however the "floor" trick falls apart when you have more than 2 ties.

Regardless, your other solution appears to be working nicely and seems to be a good deal faster than using "unique". Very nice! I'll have to look into this more on Monday morning when I have a clear head, its quitting time now.

Thanks again!

[Oleg Komarov]

floor(tiedrank... was dirty trick :)

[sunbeam]

There is something wrong here; for

data = [1 -3 -3 2 23 23];

The result is

rnk =

     3     1     1     4     5     4

[Tommaso Fornaciari]

Hi is there a way to assign equal observations two different subsequent ranks? following on the original question, the output i would need is 2 5 4 1 3 or 2 5 3 1 4

Thank you

Answer 2

[Raph]

It should also work with sort() and ismember()

data_sorted = sort(data);
[~, rnk] = ismember(data,data_sorted)

[Bradley Stiritz]

Very impressive, Raph! Thanks for your contribution. Excellent use of built-in vectorized functions.

Answer 3

[sunbeam]

This should work. I couldn't figure out how to do it without a loop, but at least this only loops over the duplicate entries. Someone let me know if you come up with a better way.

function outrank = rankWithDuplicates(data,mode)
% R = rankWithDuplicates(data,mode) ranks the values in the data variable
% according to size, allowing for duplicates. Whereas sort actually
% rearranges the input, and therefore duplicates get assigned different
% indices, rankWithDuplicates will simply output the rank order allowing
% ties for duplicate entries. For example,
%
%  rankWithDuplicates([1 1 5 8 8 10])
%
% will output [1 1 3 4 4 6]; and if these entries are shuffled like
%
%  rankWithDuplicates([8 1 5 1 10 8])
%
% the output will be [4 1 3 1 6 4].
%
% INPUT: data, a vector of real numbers.
%        mode, an optional input which can be 'ascend' or 'descend'
%
% OUTPUT: the rank order of the input data.
%

if nargin==1
    mode='ascend';
end

[~,b]=size(data);
if b==1
    data=data';
end

% Sort data
[srt, idxSrt]  = sort(data,mode);
% Find where are the repetitions and negate
idxRepeat = [false diff(srt) == 0];

% Loop through where there are duplicates and maintain the rank. 
% I'm not sure if this is necessary but it's the only way I could get it
% done.

rnk = 1:numel(data);
loopidx=find(idxRepeat>0);
for i=loopidx
    rnk(i)=rnk(i-1);
end

% Return order according to original sort

outrank(idxSrt)=rnk;

Answer 4

[Jeyamugan T]

I wrote this code for some other purpose but it may useful for this problem.

function [rkList]=arrayRankEx(O)
    cO=sort(O);
    n=size(O,2);
    rkList=zeros(1,n);
    in=1;
    while(in<=n)
        out=1;
        co=0;
        while(out<=n && in<=n)
            if(O(out)==cO(in))
                rkList(out)=in;
                co=co+1;
            end
            out=out+1;
        end 
        in=in+co;
    end
end

 >>[5 7 -2 1 -1 0 0 1 5 3]

ans =

     5     7    -2     1    -1     0     0     1     5     3

>> arrayRankEx([5 7 -2 1 -1 0 0 1 5 3])

ans =

     8    10     1     5     2     3     3     5     8     7

Answer 5

[Benjamin Levy]

Not sure if this is still a 'live' thread, but the code should report these ranks for ascending order: 2.0000 5.0000 3.5000 1.0000 3.5000.

Now, suppose your data set is data = [ 11 20 2 14 15 11 13 20 7 9 1 5 17... 7 5 16 3 5 20 ]; Your answer for ascending order (correcting for ties), using sortrows([ data' ranks ],2), should provide column 1 = data, column 2 = ranks:

    1.0000    1.0000
    2.0000    2.0000
    3.0000    3.0000
    5.0000    5.0000
    5.0000    5.0000
    5.0000    5.0000
    7.0000    7.5000
    7.0000    7.5000
    9.0000    9.0000
   11.0000   11.0000
   11.0000   11.0000
   11.0000   11.0000
   13.0000   13.0000
   14.0000   14.0000
   15.0000   15.0000
   16.0000   16.0000
   17.0000   17.0000
   20.0000   19.0000
   20.0000   19.0000
   20.0000   19.0000

Note that there are several sections in the sorted data wherein there are consecutive runs of same integers (e.g., ...5 5 7 7 ).

Using your code and my data set, and the same final sort, I have (column 1 data, column 2 ranks):

     1     1
     2     2
     3     3
     5     4
     5     4
     5     4
     7     5
    11     7
     7     7
    11     7
     9     9
    11    10
    13    13
    20    13
    20    13
    14    14
    15    15
    16    16
    17    17
    20    18

[Nataraja M]

Hello Sir I used above command sortrows([ data' ranks ],2) for ranking vectors from maximum to lowest, but facing error like Not enough input arguments. Can you please help me to solve this error Thank you