Beefy Boxes and Bandwidth Generously Provided by pair Networks
Pathologically Eclectic Rubbish Lister
 
PerlMonks  

Re: Seeking algorithm for finding common continous sub-patterns

by tachyon (Chancellor)
on Dec 03, 2004 at 09:33 UTC ( #412073=note: print w/ replies, xml ) Need Help??


in reply to Seeking algorithm for finding common continous sub-patterns

Maybe this will be fast enough for you. Maybe you need to be a bit more specific with your spec. This will not scale very well. The 27 element data set consumes 120 loops in the generation phase and 34 in the output phase for a total of 154. It is roughly O(n^2) but it is quite data dependent.

I wrote Algorithm::LCSS which is based on Algorithm::Diff and may be a better option depending on the real task. The problem with that approach is that it is a one to one comparison not many to many which is what you seem to want.

my @a = ( [2,5,10,5,12,6,21,5,10,12,23], [5,6,11,10,5,10,6,21,5,1,9], [6,5,10,15,21] ); my $m = 2; my $n = 2; my %h; for my $ref (@a) { for my $i ( 0 .. (@$ref-$n) ) { for my $j ( ($i+$n-1) .. @$ref-1 ) { $h{ join ',', (@$ref)[$i..$j] }++; } } } for my $seq( sort { $h{$b}<=>$h{$a} } keys %h ) { print "$h{$seq}: $seq\n" if $h{$seq} >= $m; } __DATA__ 4: 5,10 2: 6,21,5 2: 10,5 2: 6,21 2: 21,5

cheers

tachyon


Comment on Re: Seeking algorithm for finding common continous sub-patterns
Download Code

Log In?
Username:
Password:

What's my password?
Create A New User
Node Status?
node history
Node Type: note [id://412073]
help
Chatterbox?
and the web crawler heard nothing...

How do I use this? | Other CB clients
Other Users?
Others chanting in the Monastery: (5)
As of 2014-07-29 01:24 GMT
Sections?
Information?
Find Nodes?
Leftovers?
    Voting Booth?

    My favorite superfluous repetitious redundant duplicative phrase is:









    Results (211 votes), past polls