czyrym 1.0.3

Python library to decide if two polish words rhyme

czyrym

A Python library to detect if two Polish words rhyme. The library has been tested on a corpus of Polish rhymes from czterycztery.pl.

Performance

The corpus contains 8100 rhymes - czyrym successfully detects 7387 of them. The rate of false positives is quite low: from 37177 non-rhymes in the corpus, czyrym incorrectly detects only 239 of them as rhymes (and for some of them it can be argued that they really are rhymes, just quite imperfect).

When czyrym detects a pair of words as rhymes, it gives a distance between the two words as a number between 0 (a perfect rhyme) and a very big number (hardly a rhyme at all) - so if you want czyrym to be more conservative, you can compare this distance with your threshold.

Usage Example

import czyrym
from typing import Optional

first: str = czyrym.normalize_word('Róż, ty???')
second: str = czyrym.normalize_word('ruszty')

# if you just want to know if two words rhyme or not, do:
if czyrym.is_rhyme(first, second):
    print(f'"{first}" and "{second}" rhyme')
else:
    print(f'"{first}" and "{second}" do not rhyme')

# if you want to know if two words rhyme or not and what is the distance between them, do:
match: Optional[czyrym.RhymeMatch] = czyrym.find_rhyme_match(first, second)
if match is None:
    print(f'"{first}" and "{second}" don\'t rhyme')
else:
    print(f'A match for "{first}" and "{second}" was found - they rhyme. Total cost of the rhyme (a distance between words) is {match.total_cost}.')

# if you want to understand how it was decided that two words rhyme, do:
if match is None:
    print(f'a match for "{first}" and "{second}" was not found - they don\'t rhyme')
else:
    print(f'A match for "{first}" and "{second}" was found - they rhyme. Their common suffix is {match.common_form}.')
    for word, path in (first, match.first_path), (second, match.second_path):
        if len(path.steps) <= 1:
            print(f'For "{word}" the suffix was not mutated.')
        else:
            print(f'For "{word}" the suffix was produced with these steps:')
            for step in path.steps:
                print(f'  "{step.before}" -> "{step.after}" (mutator: {step.mutator_name}, cost: {step.cost})')
            print(f'Total cost of these steps was {path.cost}.')