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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 ( #482072=note: print w/replies, xml ) Need Help??

... 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
[download]```

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;
[download]```

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;
}
[download]```

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.
Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
"Science is about questioning the status quo. Questioning authority".
The "good enough" maybe good enough for the now, and perfection maybe unobtainable, but that should not preclude us from striving for perfection, when time, circumstance or desire allow.

Replies are listed 'Best First'.
Re^7: Algorithm for cancelling common factors between two lists of multiplicands
by tmoertel (Chaplain) on Aug 09, 2005 at 00:40 UTC
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 = []}
[download]```
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
[download]```
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)
[download]```
You didn't do anything wrong, but you did ask for the internal representation (via rpCutoff) instead of the final ratio (via pCutoff). The internal representation is likely to be insanly huge for large problems.

Save the other code as FishersExactTest.hs and the following as Main.hs and then compile:
ghc -O2 -o fet --make Main.hs
to get a program fet that you can pipe a matrix into.

```module Main (main) where
import FishersExactTest (pCutoff)
main = interact \$ show . pCutoff . map (map read . words) . lines
[download]```
See my earlier node (Re^5: Algorithm for cancelling common factors between two lists of multiplicands) for an example of usage.

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