How can I split multiple joined words?
The best tool for the job here is recursion, not regular expressions. The basic idea is to start from the beginning of the string looking for a word, then take the remainder of the string and look for another word, and so on until the end of the string is reached. A recursive solution is natural since backtracking needs to happen when a given remainder of the string cannot be broken into a set of words. The solution below uses a dictionary to determine what is a word and prints out solutions as it finds them (some strings can be broken out into multiple possible sets of words, for example wickedweather could be parsed as "wicked we at her"). If you just want one set of words you will need to determine the rules for selecting the best set, perhaps by selecting the solution with fewest number of words or by setting a minimum word length.
#!/usr/bin/perl
use strict;
my $WORD_FILE = '/usr/share/dict/words'; #Change as needed
my %words; # Hash of words in dictionary
# Open dictionary, load words into hash
open(WORDS, $WORD_FILE) or die "Failed to open dictionary: $!\n";
while (<WORDS>) {
chomp;
$words{lc($_)} = 1;
}
close(WORDS);
# Read one line at a time from stdin, break into words
while (<>) {
chomp;
my @words;
find_words(lc($_));
}
sub find_words {
# Print every way $string can be parsed into whole words
my $string = shift;
my @words = @_;
my $length = length $string;
foreach my $i ( 1 .. $length ) {
my $word = substr $string, 0, $i;
my $remainder = substr $string, $i, $length - $i;
# Some dictionaries contain each letter as a word
next if ($i == 1 && ($word ne "a" && $word ne "i"));
if (defined($words{$word})) {
push @words, $word;
if ($remainder eq "") {
print join(' ', @words), "\n";
return;
} else {
find_words($remainder, @words);
}
pop @words;
}
}
return;
}
Can a human do it?
farsidebag far sidebag farside bag far side bag
Not only do you have to use a dictionary, you might have to use a statistical approach to figure out what's most likely (or, god forbid, an actual HMM for your human language of choice...)
For how to do statistics that might be helpful, I turn you to Dr. Peter Norvig, who addresses a different, but related problem of spell-checking in 21 lines of code: http://norvig.com/spell-correct.html
(he does cheat a bit by folding every for loop into a single line.. but still).
Update This got stuck in my head, so I had to birth it today. This code does a similar split to the one described by Robert Gamble, but then it orders the results based on word frequency in the provided dictionary file (which is now expected to be some text representative of your domain or English in general. I used big.txt from Norvig, linked above, and catted a dictionary to it, to cover missing words).
A combination of two words will most of the time beat a combination of 3 words, unless the frequency difference is enormous.
I posted this code with some minor changes on my blog
http://squarecog.wordpress.com/2008/10/19/splitting-words-joined-into-a-single-string/ and also wrote a little about the underflow bug in this code.. I was tempted to just quietly fix it, but figured this may help some folks who haven't seen the log trick before: http://squarecog.wordpress.com/2009/01/10/dealing-with-underflow-in-joint-probability-calculations/
Output on your words, plus a few of my own -- notice what happens with "orcore":
perl splitwords.pl big.txt words answerveal: 2 possibilities - answer veal - answer ve al wickedweather: 4 possibilities - wicked weather - wicked we at her - wick ed weather - wick ed we at her liquidweather: 6 possibilities - liquid weather - liquid we at her - li quid weather - li quid we at her - li qu id weather - li qu id we at her driveourtrucks: 1 possibilities - drive our trucks gocompact: 1 possibilities - go compact slimprojector: 2 possibilities - slim projector - slim project or orcore: 3 possibilities - or core - or co re - orc ore
Code:
#!/usr/bin/env perl
use strict;
use warnings;
sub find_matches($);
sub find_matches_rec($\@\@);
sub find_word_seq_score(@);
sub get_word_stats($);
sub print_results($@);
sub Usage();
our(%DICT,$TOTAL);
{
my( $dict_file, $word_file ) = @ARGV;
($dict_file && $word_file) or die(Usage);
{
my $DICT;
($DICT, $TOTAL) = get_word_stats($dict_file);
%DICT = %$DICT;
}
{
open( my $WORDS, '<', $word_file ) or die "unable to open $word_file\n";
foreach my $word (<$WORDS>) {
chomp $word;
my $arr = find_matches($word);
local $_;
# Schwartzian Transform
my @sorted_arr =
map { $_->[0] }
sort { $b->[1] <=> $a->[1] }
map {
[ $_, find_word_seq_score(@$_) ]
}
@$arr;
print_results( $word, @sorted_arr );
}
close $WORDS;
}
}
sub find_matches($){
my( $string ) = @_;
my @found_parses;
my @words;
find_matches_rec( $string, @words, @found_parses );
return @found_parses if wantarray;
return \@found_parses;
}
sub find_matches_rec($\@\@){
my( $string, $words_sofar, $found_parses ) = @_;
my $length = length $string;
unless( $length ){
push @$found_parses, $words_sofar;
return @$found_parses if wantarray;
return $found_parses;
}
foreach my $i ( 2..$length ){
my $prefix = substr($string, 0, $i);
my $suffix = substr($string, $i, $length-$i);
if( exists $DICT{$prefix} ){
my @words = ( @$words_sofar, $prefix );
find_matches_rec( $suffix, @words, @$found_parses );
}
}
return @$found_parses if wantarray;
return $found_parses;
}
## Just a simple joint probability
## assumes independence between words, which is obviously untrue
## that's why this is broken out -- feel free to add better brains
sub find_word_seq_score(@){
my( @words ) = @_;
local $_;
my $score = 1;
foreach ( @words ){
$score = $score * $DICT{$_} / $TOTAL;
}
return $score;
}
sub get_word_stats($){
my ($filename) = @_;
open(my $DICT, '<', $filename) or die "unable to open $filename\n";
local $/= undef;
local $_;
my %dict;
my $total = 0;
while ( <$DICT> ){
foreach ( split(/\b/, $_) ) {
$dict{$_} += 1;
$total++;
}
}
close $DICT;
return (\%dict, $total);
}
sub print_results($@){
#( 'word', [qw'test one'], [qw'test two'], ... )
my ($word, @combos) = @_;
local $_;
my $possible = scalar @combos;
print "$word: $possible possibilities\n";
foreach (@combos) {
print ' - ', join(' ', @$_), "\n";
}
print "\n";
}
sub Usage(){
return "$0 /path/to/dictionary /path/to/your_words";
}
The Viterbi algorithm is much faster. It computes the same scores as the recursive search in Dmitry's answer above, but in O(n) time. (Dmitry's search takes exponential time; Viterbi does it by dynamic programming.)
import re
from collections import Counter
def viterbi_segment(text):
probs, lasts = [1.0], [0]
for i in range(1, len(text) + 1):
prob_k, k = max((probs[j] * word_prob(text[j:i]), j)
for j in range(max(0, i - max_word_length), i))
probs.append(prob_k)
lasts.append(k)
words = []
i = len(text)
while 0 < i:
words.append(text[lasts[i]:i])
i = lasts[i]
words.reverse()
return words, probs[-1]
def word_prob(word): return dictionary[word] / total
def words(text): return re.findall('[a-z]+', text.lower())
dictionary = Counter(words(open('big.txt').read()))
max_word_length = max(map(len, dictionary))
total = float(sum(dictionary.values()))
Testing it:
>>> viterbi_segment('wickedweather')
(['wicked', 'weather'], 5.1518198982768158e-10)
>>> ' '.join(viterbi_segment('itseasyformetosplitlongruntogetherblocks')[0])
'its easy for me to split long run together blocks'
To be practical you'll likely want a couple refinements:
- Add logs of probabilities, don't multiply probabilities. This avoids floating-point underflow.
- Your inputs will in general use words not in your corpus. These substrings must be assigned a nonzero probability as words, or you end up with no solution or a bad solution. (That's just as true for the above exponential search algorithm.) This probability has to be siphoned off the corpus words' probabilities and distributed plausibly among all other word candidates: the general topic is known as smoothing in statistical language models. (You can get away with some pretty rough hacks, though.) This is where the O(n) Viterbi algorithm blows away the search algorithm, because considering non-corpus words blows up the branching factor.
pip install wordninja
>>> import wordninja
>>> wordninja.split('bettergood')
['better', 'good']