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Re^5: Algorithm for cancelling common factors between two lists of multiplicands

by tmoertel (Chaplain)
 on Aug 08, 2005 at 23:31 UTC ( #482064=note: print w/replies, xml ) Need Help??

Quick question: If the sample size is so large, is there a reason you aren't using the Chi-square test? My understanding is Fisher's Exact Test may be preferred when the sample size isn't large enough to reasonably support the large-sample approximation for the Chi-square test. Since you have a large sample, why not take the easy road?
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Re^6: Algorithm for cancelling common factors between two lists of multiplicands
by BrowserUk (Pope) on Aug 09, 2005 at 00:03 UTC
... is there a reason you aren't using the Chi-square test?

I'm not trying to solve the sample problem. That was just an example I found on the web and used as a test.

I set out to solve the problem of performing the FET using Perl. First I did it with Math::Pari, but which gives a result of 8.070604647867604097576877668E-7030 in 26ms, but I was unsure about the accuracy. It also imposes a binary dependancy.

#! perl -slw use strict; use Benchmark::Timer; use List::Util qw[ sum reduce ]; use Math::Pari qw[ factorial ]; \$a=\$b; sub product{ reduce{ \$a *= \$b } 1, @_ } sub FishersExactTest { my @data = @_; return unless @data == 4; my @C = ( sum( @data[ 0, 2 ] ), sum( @data[ 1, 3 ] ) ); my @R = ( sum( @data[ 0, 1 ] ), sum( @data[ 2, 3 ] ) ); my \$N = sum @C; my \$dividend = product map{ factorial \$_ } grep \$_, @R, @C; my \$divisor = product map{ factorial \$_ } grep \$_, \$N, @data; return \$dividend / \$divisor; } my \$T = new Benchmark::Timer; \$T->start( '' ); print FishersExactTest 989, 9400, 43300, 2400;; \$T->stop( '' ); \$T->report; __END__ P:\test>MP-FET.pl 8.070604647867604097576877668E-7030 1 trial of _default ( 25.852ms total), 25.852ms/trial

So, then I coded it using Math::BigFloat

#! perl -slw use strict; use Benchmark::Timer; use List::Util qw[ reduce ]; use Math::BigFloat; \$a=\$b; sub product{ reduce{ \$a *= \$b } 1, @_ } sub sum{ reduce{ \$a += \$b } 0, @_ } sub FishersExactTest { my @data = map{ Math::BigFloat->new( \$_ ) } @_; return unless @data == 4; my @C = ( sum( @data[ 0, 2 ] ), sum( @data[ 1, 3 ] ) ); my @R = ( sum( @data[ 0, 1 ] ), sum( @data[ 2, 3 ] ) ); my \$N = sum @C; my \$dividend = product map{ \$_->bfac } grep \$_, @R, @C; my \$divisor = product map{ \$_->bfac } grep \$_, \$N, @data; return \$dividend / \$divisor; } my \$T = new Benchmark::Timer; \$T->start( '' ); print FishersExactTest 989, 9400, 43300, 2400;; \$T->stop( '' ); \$T->report;

But that ran for 20 minutes without producing any output before I killed it (I've set it running again now, and my machines fan has been thrashing at full speed for the last 25 minutes).

Whilst I was waiting for the BigFloat version, I coded this version which attempts to reduce the size of the problem by eliminating (exactly common) factors:

sub FishersExactTest2 { my @data = @_; return unless @data == 4; my @C = ( sum( @data[ 0, 2 ] ), sum( @data[ 1, 3 ] ) ); my @R = ( sum( @data[ 0, 1 ] ), sum( @data[ 2, 3 ] ) ); my \$N = sum @C; my %dividends; \$dividends{ \$_ }++ for map{ factors \$_ } grep \$_, @ +R, @C; my %divisors; \$divisors { \$_ }++ for map{ factors \$_ } grep \$_, \$ +N, @data; for my \$i ( keys %divisors ) { if( exists \$dividends{ \$i } ) { \$divisors{ \$i }--, \$dividends{ \$i }-- while \$divisors{ \$i } and \$dividends{ \$i }; delete \$divisors { \$i } unless \$divisors { \$i }; delete \$dividends{ \$i } unless \$dividends{ \$i }; } } my \$dividend = product( map{ ( \$_ ) x \$dividends{ \$_ } } keys %div +idends ); my \$divisor = product( map{ ( \$_ ) x \$divisors { \$_ } } keys %div +isors ); return \$dividend / \$divisor; }

This works well for values smallish values, but cannot handle the example I gave above (NV overflow).

It was then I started thinking about how to eliminate more factors from the equation so as to reduce the size of the intermediate terms, and posted my SoPW. I think that hv's solution of expanding all terms to their prime factorizations before performing the cancelling out will be a winner--but I haven't finished coding that yet.

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
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My Haskell implementation represents numbers as the ratio of products of ordered integer streams. For example, I represent 3!/(4*5) as (R numerator=[1,2,3] denominator=[4,5]). In this representation, multiplication becomes merging the numerator and denominator streams and then canceling the first stream by the second. In this way I can remove all cancelable original terms in the Pcutoff formula before finally multiplying the terms that remain.
*FishersExactTest> fac 6 R {numer = [2,3,4,5,6], denom = []} *FishersExactTest> fac 3 R {numer = [2,3], denom = []} *FishersExactTest> fac 6 `rdivide` fac 3 R {numer = [4,5,6], denom = []}
Here's the example from the MathWorld page:
*FishersExactTest> rpCutoff [ [5,0], [1,4] ] R {numer = [2,3,4,5], denom = [7,8,9,10]} *FishersExactTest> fromRational . toRatio \$ it 2.3809523809523808e-2
The code:
module FishersExactTest (pCutoff) where import Data.Ratio import Data.List (transpose) pCutoff = toRatio . rpCutoff rpCutoff rows = facproduct (rs ++ cs) `rdivide` facproduct (n:xs) where rs = map sum rows cs = map sum (transpose rows) n = sum rs xs = concat rows -- cells facproduct = rproduct . map fac fac n | n < 2 = runit | otherwise = R [2..n] [] -- I represent numbers as ratios of products of integer streams -- R [1,2,3] [4,5] === (1 * 2 * 3) / (4 * 5) data Rops = R { numer :: [Int], denom :: [Int] } deriving Show runit = R [] [] -- the number 1 toRatio (R ns ds) = bigProduct ns % bigProduct ds bigProduct = product . map toInteger -- multiplication is merging numerator and denominator streams -- and then canceling the first by the second rtimes (R xns xds) (R yns yds) = uncurry R \$ (merge xns yns) `cancel` (merge xds yds) rproduct = foldr rtimes runit -- division is multiplication by the inverse rdivide x (R yns yds) = rtimes x (R yds yns) -- helpers merge (x:xs) (y:ys) | x < y = x : merge xs (y:ys) | otherwise = y : merge (x:xs) ys merge [] ys = ys merge xs [] = xs cancel (x:xs) (y:ys) | x == y = cancel xs ys | x < y = let (xs', ys') = cancel xs (y:ys) in (x:xs', ys') | otherwise = let (xs', ys') = cancel (x:xs) ys in (xs', y:ys') cancel xs ys = (xs, ys)

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