Automatically trimming a monodix
At the moment we have a problem in Apertium regarding copies. When we come to make a new language pair that is based off some existing resource in Apertium (e.g. a monodix), then usually what we do is make a copy of that resource, and then change it as we need. This is less than ideal because:
- it means that any improvements we make aren't automatically carried over to the dictionary we copied from
- it means that any improvements in the dictionary we copied from aren't carried over into our new dictionary.
So why do we do it ? -- Testvoc. If we have entries in our monodix that aren't in our bidix, then we get lots of @
in our output. This is bad.
One way to get around the problem is to use ad hoc scripts for trimming the dictionaries (see for example: apertium-af-nl, apertium-sme-nob and the trim-lexc.py
script in trunk/apertium-tools
), but these are less than ideal because usually they have to include specific hacks for differences in dictionary format.
Another way is to take the intersection of our monodix and our bidix, and use that for analysis. That's what is described below.
Note: This doesn't take care of the whole testvoc problem. It would still be necessary to get rid of #
symbols.
Contents
Example
Suppose you have the monodix:
<dictionary> <alphabet>abcdefghijklmnopqrstuvwxyz</alphabet> <sdefs> <sdef n="n"/> <sdef n="sg"/> <sdef n="pl"/> </sdefs> <pardefs> <pardef n="beer__n"> <e><p><l></l><r><s n="n"/><s n="sg"/></r></p></e> <e><p><l>s</l><r><s n="n"/><s n="pl"/></r></p></e> </pardef> </pardefs> <section id="main" type="standard"> <e lm="beer"><i>beer</i><par n="beer__n"/></e> <e lm="school"><i>school</i><par n="beer__n"/></e> <e lm="computer"><i>computer</i><par n="beer__n"/></e> <e lm="house"><i>house</i><par n="beer__n"/></e> </section> </dictionary>
It generates the following strings:
$ lt-expand test-en.dix beer:beer<n><sg> beers:beer<n><pl> school:school<n><sg> schools:school<n><pl> computer:computer<n><sg> computers:computer<n><pl> house:house<n><sg> houses:house<n><pl>
But our bidix is only:
<dictionary> <alphabet>abcdefghijklmnopqrstuvwxyz</alphabet> <sdefs> <sdef n="n"/> <sdef n="sg"/> <sdef n="pl"/> </sdefs> <pardefs> </pardefs> <section id="main" type="standard"> <e><p><l>beer<s n="n"/></l><r>garagardo<s n="n"/></r></p></e> <e><p><l>house<s n="n"/></l><r>etxe<s n="n"/></r></p></e> </section> </dictionary>
We don't want to include the entries for "computer" and "school", because then we would get @
in our output.
HFST
Here is a Makefile that given two dictionaries test-en.dix
(the monodix) and test-en-eu.dix
(the bidix), will produce a binary transducer of the monodix in HFST format, that only contains the strings matching prefixes in the bidix.
all: lt-comp lr test-en.dix test-en.bin lt-comp lr test-en-eu.dix test-en-eu.bin lt-print test-en.bin > test-en.att lt-print test-en-eu.bin > test-en-eu.att hfst-txt2fst -e ε < test-en.att > test-en.fst hfst-txt2fst -e ε < test-en-eu.att > test-en-eu.fst hfst-invert test-en.fst -o test-en.mor.fst hfst-project -p upper test-en-eu.fst > test-en-eu.en.fst echo " ?* " | hfst-regexp2fst > any.fst hfst-concatenate -1 test-en-eu.en.fst -2 any.fst -o test-en-eu.en-prefixes.fst hfst-compose-intersect -1 test-en-eu.en-prefixes.fst -2 test-en.mor.fst | hfst-invert -o test-en.trimmed.fst clean: rm *.bin *.att *.fst
If we run hfst-fst2strings
, we get:
$ hfst-fst2strings test-en.trimmed.fst beer:beer<n><sg> beers:beer<n><pl> house:house<n><sg> houses:house<n><pl>
lttoolbox
$ lt-comp rl apertium-en-ca.en-ca.dix ca-en.autobil.bin $ lt-comp lr apertium-en-ca.ca.dix ca-en.automorf-full.bin $ lt-trim ca-en.automorf-full.bin ca-en.autobil.bin ca-en.automorf.bin
Where the left side of the second transducer (ca-en.autobil.bin) is converted into prefixes, and only strings from the first transducer which match the prefixes in the second transducer are included into the final compilation.