Converting BCBCBCBCBCCCB to (BC)^{5}CCB

fishfork has asked for the wisdom of the Perl Monks concerning the following question:

Hi,

A mathematical program I wrote in C++ multiplies things called B and C together then finds some properties of the new thing.

The output gives the thing, for example
BCBCBCBCBCCCB
then a list of porperties.

I am now writing a perl script to process this output.

I want to replace
BCBCBCBCBCCCB
by
(BC)^{5}CCB
which is obviously a lot nicer.

However, I some of the 'words' start with C and the number of (BC)s ranges from 0 to some very large numbers.

How do I achieve this? I tried a few ideas but none of them worked and now I really don't have a clue what to do!

Thanks for reading,
Richard.
richard@sigma.ndo.co.uk

Edit by tye to preserve formatting

Comment on Converting BCBCBCBCBCCCB to (BC)^{5}CCB

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Re: Converting BCBCBCBCBCCCB to (BC)^{5}CCB by gmax (Abbot) on Feb 06, 2002 at 14:54 UTC
If the case is that simple (your string could start by /^BC/ or by /^.*BC/), you can use the replacement operator s/// to get the right answer. `#!/usr/bin/perl -w use strict; my $source = "BCBCBCBCBCCCB"; my $group = "BC"; # the repeating string # capturing the left part, if any, # and removing it from the source my $left = substr($source,0, index($source,$group)); substr($source,0,index($source,$group)) =""; # counting the occurrence of the repeating group my $exp = $source =~ s/$group//g; #print the "formula" print "$left ($group)^{$exp}$source\n";` [download] Output: `(BC)^{5}CCB` Changing source to "CCBCBCBCBCBCCCB"; You get the output: `CC (BC)^{5}CCB`	[reply] [d/l] [select]
Re: Converting BCBCBCBCBCCCB to (BC)^{5}CCB by talexb (Chancellor) on Feb 06, 2002 at 15:42 UTC
I'd use a regexp to capture the longest string with repeated `BC` elements, then divide the length of that string by the length of the original `BC` string to get the exponent, and finish by deleting the stuff just exponentiated. `$Code = "BCBCBCBCBCCCB"; $What = "BC"; $Code =~ m/($What)+?/; $Exp = length ( $1 ) / length ( $What ); $Code =~ s/$1/($What)^($Exp)/e;` [download] This code not tested .. that's just the general idea. --t. alex "Of course, you realize that this means war." -- Bugs Bunny.	[reply] [d/l] [select]
Re: Converting BCBCBCBCBCCCB to (BC)^{5}CCB by particle (Vicar) on Feb 06, 2002 at 17:26 UTC
some good advice has been given above. if performance is of any concern, you'll save a little time by compiling the regex only once, by using the -o modifier, such as `my $exp = $source =~ s/$group//go;` to borrow gmax's solution for illustration. of course, direct interpolation of the pattern will be even faster, such as `my $exp = $source =~ s/BC//g;` my benchmark results look like: `C:\WINDOWS\Desktop>perl test_regex-o.pl Benchmark: timing 100000 iterations of s_direct, s_witho, s_withouto.. +. s_direct: 4 wallclock secs ( 3.02 usr + 0.00 sys = 3.02 CPU) @ 33 +112.58/s (n=100000) s_witho: 4 wallclock secs ( 3.41 usr + 0.00 sys = 3.41 CPU) @ 29 +325.51/s (n=100000) s_withouto: 5 wallclock secs ( 4.28 usr + 0.00 sys = 4.28 CPU) @ 23 +364.49/s (n=100000)` [download] ~Particle	[reply] [d/l] [select]
Re: Converting BCBCBCBCBCCCB to (BC)^{5}CCB by belg4mit (Prior) on Feb 06, 2002 at 14:37 UTC
This is runlength encoding(RLE), searching the web for such might help. There is mention of a perl module `Marshal::Packed` (by MUIR) which does RLE, but it is does not appear to be available anywhere.	[reply]