bestthing.info algorithm

readme.md

This is a simple algorithm as described by bestthings.info. An entity's rank is calculated by (Num_things_I_beat)/(Num_things_I_beat + Num_things_beat_me)

I put this together in about half an hour (I'm really starving for time right now, with tons of projects!). I have a few ideas for the second part, but alas no code.

Part 2) We will aim to improve the accuracy of our ranking by trying to re-rank those pairs that are not different enough. For example, if the agg between a pair is say < 10, they are too similar, and we would wish to re-rank them. This could be done by implementing a simple dictionary to store aggregate ranks..

aggregate = {('a','b'): 23, ('a','c'): 3}

Now we could scan this dict to present to a user those pairs that are not different enough. For those with very high aggregate scores, we can avoid re-ranking because "the voters have had their say."

findbest.py

#!/usr/bin/env python

import cPickle as pickle
import random

# things I beat
tIB = {}
# things beat me
tBM = {}

numObj = random.randint(10,100)

corpus = {}

# Constructing the corpus
for i in range(numObj):
	name = str(i)
	corpus[name] = {}
	for j in range(numObj):
		if(j!=i):
			opp = str(j)
			corpus[name][opp] = random.randint(20,5000)

# picklable for portability
pickle.dump( corpus, open( "best.p", "wb" ) )

corpus = pickle.load(open("best.p","rb"))

# find who beat whom.
for k in corpus:
	numObj += 1
	tIB[k] = 0
	tBM[k] = 0
	for opp in corpus[k]:
		myScore = corpus[k][opp]
		oppScore = corpus[opp][k]
		agg = myScore - oppScore
		if(agg > 0):
			tIB[k]+=1
		elif(agg < 0):
			tBM[k]+=1
		else:
			tIB[k]+=0.5
			tBM[k]+=0.5

rank = {}

# Assign ranks
for x in tIB:
	rank[x] =  (float(tIB[x])/(float(tIB[x]+tBM[x])))

# Sort and print
for key,value in sorted(rank.iteritems(), key=lambda (k,v): (v,k)):
    print "%s: %s" % (key, value)