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LURE: sk

Scott Knott test

tim@menzies.us
August'17

Recursive bi-clustering of set of samples. Before desceding into either split, pause and check if the splits are statistically different. Returns the samples, sorted on their median, labelled with a rank number. If two adjecent samples are statistically the same, then they get the same rank. If all samples get ranked "1" then that means there are no interesting distinctions between these samples.

 
require "show"
local the=require "config"
local tiles=require "tiles"
local lst=require "lists"

Simple utils

 
local function nth(t,n) return t._all[  math.floor(#t._all*n) ] end
local function mid(t)   return nth(t,0.5) end
local function iqr(t)   return nth(t,0.75) - nth(t,0.25) end

Here's a simple counter, used to track mu and all.

 
local function create() return {
  _all={}, sum=0,n=0, mu=0} end

Here how to update a counter with one value x.

 
local function update(i,x) 
  i._all[#i._all+1]=x
  i.sum = i.sum + x
  i.n   = i.n + 1 
  i.mu  = i.sum/i.n end

Here how to update a counter with many values from t.

 
local function updates(t, counter)
  counter = counter or create()
  for j=1,#t do 
    update(counter,t[j]) end 
  return counter end

This code is always counting left and right within the list of samples. To save time, memo all those "peeks".

 
local function memo(samples,here,stop,_memo,    b4,inc)
  if stop > here then inc=1 else inc=-1 end
  if here ~= stop then  
     b4=  lst.copy( memo(samples,here+inc, stop, _memo)) end
  _memo[here] = updates(samples[here]._all,  b4)
  return _memo[here] end

Seek a split that maximizes the expected value of the square of the difference in means before and after the split. At that point, if the two splits are not statistically the same, then recurse into each part of the split.

 
return function (samples,epsilon,same)
  epsilon = epsilon or the.sample.epsilon
    local function split(lo,hi,all,rank,lvl)   
      local best,lmemo,rmemo = 0,{},{}
      memo(samples,hi,lo, lmemo) -- summarize i+1 using i
      memo(samples,lo,hi, rmemo) -- summarize i using i+1
      local cut, lbest, rbest
      for j=lo,hi-1 do -- step1: look for the best cut
        local l = lmemo[j]
        local r = rmemo[j+1]
        if mid(l)*the.sample.epsilon < mid(r) then
          if not same(l,r) then
            local tmp= l.n/all.n*(l.mu - all.mu)^2 + 
                       r.n/all.n*(r.mu - all.mu)^2
            if tmp > best then
              cut   = j
              best  = tmp
              lbest = lst.copy(l)
              rbest = lst.copy(r) end end end end
      if cut then -- step2a: use the cut (if you found it)
        rank = split(lo,   cut, lbest, rank, lvl+1) + 1
        rank = split(cut+1, hi, rbest, rank, lvl+1)
      else -- step2b: otherwise, all samples get same rank
        for j=lo,hi do
          samples[j].rank = rank end end
      return rank 
    end 
  table.sort(samples, function (x,y) return 
             mid(x) < mid(y) end)
  split(1,#samples, memo(samples,1,#samples,{}),1,0)  
  return samples end

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LURE, Copyright (c) 2017, Tim Menzies All rights reserved, BSD 3-Clause License

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