Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Prehistory

I am trying to make some statistical analysis of some experimental data, arises from measurements made ​​on an ordinal scale.

I faced with the problem of rank aggregation: to get from many "individuals" orderings (on the same set of objects) one "collective" ordering.

The most "natural" approach to this problem is Kemeny-Young method (better look primary source).

Surprisingly I found out that there is no program application for this method!!! (There is one C++ code, but it does not allow weak orderings, i.e., does not allow cases when several objects share same position at ordering).

Previously I asked some points (one, two) for constructing needed code, but now I have decide to tell all at once — because the problem is more complicated than I thought at first, and I can miss some points since I am null in Mathematica and programming.

Description

Let $R_1, R_2, R_3, ..., R_N$ denote "individuals" weak orderings of $N$ given objects $a, b, c, ...$.

Let consider $R$ as set of ordered pairs of objects. Then $(a, b)\in R$ has the usual interpretation: "$a$ is ordered at least as good as b at $R$". In case $(b, a)$ is also in $R$ we say that "both are ordered equally good". Where in case $(b,a)$ is not in $R$ we say that "$a$ is ordered above $b$" or "$b$ is ordered below $a$".

(After introducing metric we will see that it is possible to neglect — since they don't make further difference — such pairs as $(a, a), (b, b), (c, c)...$.)

To demonstrate orderings notation, here are all possible orderings for $N=3$:

$\ R_1=\{(a, b), (a, c), (b, c)\}$

$\ R_2=\{(a, c), (a, b), (c, b)\}$

$\ R_3=\{(a, b), (a, c), (b, c), (c, b)\}$

$\ R_4=\{(b, a), (b, c), (a, c)\}$

$\ R_5=\{(b, c), (b, a), (c, a)\}$

$\ R_6=\{(b, a), (b, c), (a, c), (c, a)\}$

$\ R_7=\{(c, a), (c, b), (a, b)\}$

$\ R_8=\{(c, b), (c, a), (b, a)\}$

$\ R_9=\{(c, a), (c, b), (a, b), (b, a)\}$

$R_{10}=\{(a, b), (b, a), (a, c), (b, c)\}$

$R_{11}=\{(a, c), (c, a), (a, b), (c, b)\}$

$R_{12}=\{(b, c), (c, b), (b, a), (c, a)\}$

$R_{13}=\{(a, b), (b, a), (a, c), (c, a), (b, c), (c, b)\}$

For such notation we may introduce metric $\delta$ (so-called Kemeny distance) between any two orderings $R_1$ and $R_2$ by next way: $$\delta(R_1,R_2)=|R_1\setminus R_2|+|R_2\setminus R_1|$$ where $\setminus$ means set-theoretic difference; and $||$ means cardinality of set (i.e., number of elements, because sets are finite).

The required "collective" ordering $Ω$ (so-called Kemeny mean) is such ordering of given objects, that minimizes the sum of the squares of Kemeny distances to all "individuals" orderings, i.e.: $$Ω=\min \sum_{i=1}^{N } \delta(R_i,Ω)^2$$

In fact, $Ω$ may be not unique, there may be several $Ω_1, Ω_2, ...$ for which the appropriate sums of the squares of Kemeny distances are equal and minimal. So in general case $Ω$ is set of such $Ω_i$.

How to attack

So we have several of $R_i$ as input and the goal is to output $Ω$.

It looks, like there should be brute-force with one-by-one generating of possible ordering, calculating the sum of the squares of Kemeny distances to all input orderings and verifiaction for minimum.

Even for small $N$ number of possible orderings is huge (for my data case $N=10$, there are over $10^8$ possible orderings), so we should keep less data.

share|improve this question
4  
Your statement "since I am null in Mathematica and programming." makes things somewhat complicated. The participants in this site offer their time (and sometimes knowledge) to help others in Mathematica related issues. That's sport. But it isn't a free service to solve problems, irrespective of how interesting the matter could be. There are a few exceptions to this rule, but it isn't fair to expect for free the solution to a problem that could be considered as business by a consultant (or as a paper for someone in the academy) –  belisarius Nov 27 '13 at 3:59
    
You might want to ask first whether there is a consensus ordering. Look at the eigenvalues of the matrix of Kendall tau-A coefficients for pairs of "individuals". If the first eigenvalue is not substantially larger than the second then there is no consensus. –  Ray Koopman Nov 27 '13 at 4:17
    
@belisarius I assure you, that if you knew how much time I spent trying to write this code, then you would not be angry at me. (I tried hard at Delphi 7, but Mathematica looks more suitable.) But I understand your indignation! Perhaps you could see my merit at finding reformulation: the description above is much "closer" to program code, than primary sources gives. For example, Kemeny-Snell book use matrix notation, which looks less adapted for programming. Also I hope that the code would be useful for many people. –  aeiklmkv Nov 27 '13 at 7:26
    
@RayKoopman Could you please give any references to clarify what does "substantially large" mean? By the way, I know that Kendall's W (Kendall's coefficient of concordance) is also useful for test on consensus. –  aeiklmkv Nov 27 '13 at 7:42
2  
@aeiklmkv I'm not angry, since the question looks very interesting, and there is a clear effort on your part. But that effort wasn't focused on the Mathematica part of the problem. BTW, Mathematica is better than most programming environments when it comes to manage matrices (doing matrix arithmetic, finding eigenvalues, inverting, calc determinants, solving linear systems, finding orthogonal bases whatever), so perhaps you should consider the matrix notation for the problem if you're trying it in Mathematica. –  belisarius Nov 27 '13 at 11:35
show 2 more comments

1 Answer

up vote 3 down vote accepted

[Warning: There could be issues with the averaging procedure. At least one should pay attention to allPossibleOrderings; its definition probably needs to be improved. See other two *edit*s in text below.]

I'm probably doing something nasty here but if RAM is really not a problem for you, then… here you go, a blunt brute-force approach:

KemenyDistance =
Plus @@
Composition[Length, Complement] @@@
Permutations[{##}, {2}] &;

FindKemenyMean[setOfOrderings:{__List}, allObjects:_List:{}] :=
Block[{ weakAndNonWeakOrderings, allPossibleOrderings
      , kemenySumOfSquares
      , minFound, data, \[CapitalOmega]candidate },

weakAndNonWeakOrderings@objects_List :=
With[{orderingFromOrderedList = Partition[#, 2, 1]&},
Block[{nonWeakOrderingsOnly, addWeakOrderings},
  nonWeakOrderingsOnly =
  orderingFromOrderedList /@ Permutations[#, {Length@#}]&;

  addWeakOrderings@nonWeakOrdering_ :=
  Join[#, nonWeakOrdering]& /@
  (Permutations[Reverse/@#, Length@#]&@nonWeakOrdering);

  Flatten[addWeakOrderings /@ nonWeakOrderingsOnly@objects, 1]]];

allPossibleOrderings =
weakAndNonWeakOrderings @
If[allObjects === {}
,  Composition[DeleteDuplicates, Flatten]@setOfOrderings
,  allObjects];

kemenySumOfSquares[\[CapitalOmega]_, orderingsToAverage_List] :=
Plus @@ (KemenyDistance[#, \[CapitalOmega]]^2 & /@ orderingsToAverage);

Fold[
  With[
    { currentMin = First@Cases[#1, minFound@n_ :> n, {1}, 1]
    , currentOrdering = #2 }
  , With[
     { currentSum =
       kemenySumOfSquares[currentOrdering, setOfOrderings] }
    , If[currentSum < currentMin
      ,  data[minFound@currentSum, \[CapitalOmega]candidate@currentOrdering]
      ,  #1]]] &
  , data[minFound[\[Infinity]], \[CapitalOmega]candidate[]]
  , allPossibleOrderings]]

These two definitions give you KemenyDistance and FindKemenyMean precisely in the form you described. FindKemenyMean fetches initial objects from the orderings list which will always work if your orderings are linear (= complete).

In[1]:= FindKemenyMean[
          { {{a, b}, {b, c}, {c, b}, {c, d}},
            {{b, d}, {d, c}, {c, a}},
            {{c, a}, {a, d}, {d, b}}  }]
Out[1]= data[minFound[45], \[CapitalOmega]candidate[{{b, c}, {c, a}, {a, d}}]]

You can provide list of all objects independently as its second argument but I don't guarantee anything for incomplete orderings (just don't have enough mathematical intuition for that).

Notes:

1) My calculation of all orderings, with weak ones included, for $N = 3$, evaluates list that is somewhat longer than the one you provided. I probably generate equivalent (hence redundant) weak orderings! [edit: Or maybe I should have converted them to some canonical form.]

2) The $R_i$ data structure for orderings is obviously redundant, hence definitely inefficient. [edit: not so if you don't presume that the relation is transitive, or if one must get to some kind of “complete form” of all $R_i$'s in order to calculate Kemeny mean w.r.t. them] I don't understand a thing in statistics but I suggest describing orderings in a more concise manner; Mma has gives lots of possibilites to do that. For example, you could use weighted sums, like $a + 2b + 3c$, to describe that $a < b < c$ (or maybe $a + (1+\varepsilon)b + (1+\varepsilon)^2c$ would be a better idea). Calculation of average ordering then would become an immensely more simple and efficient procedure (you could just use arithmetic mean, or something close to it), although I can't say much about its sanity. At least orderings that treat $a$ and $b$ as equal would always get averaged to orderings with the same property in case you pick simple arithmetic mean.

3) I use a lot of nested scoping constructs. Normally, one would write a small package instead. I hope this won't make it significantly more dificult for you to find out what's going on here in case you're interested in Mathematica's core functions. Hopefully, explicit names of symbols (I could say it's a [really good] Mma tradition) will be of help.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.