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Wordplay

Build Status PyPI version Supported versions PyPI version

Python package for word searching utilities.

Using a convenient API, you can filter a set of strings with detailed criteria. There are also more scrabble-like features such as getting all anagrams of a word.

Install

pip install wordplay

Background

I initially built this as an algorithm solely to help my endeavors in word games such as Scrabble. It was useful for finding word combinations in every type of situation. However, I realized it could be used for many other purposes, such as filtering email addresses, phone numbers and any set of data really. So I decided to decouple the API from my personal use.

Usage

Words are stored in a Dictionary object. Initialize the dictionary with a set of strings (there are no restrictions on what the string can contain) or Dictionary() with no arguments to use the word set sourced from here. In the future, I will amass a couple of word lists and make them options for initializing the dictionary.

If you do want restrictions on the string, see the documentation for the Utils module. For complex query parameters, you can use a Criteria object. The class uses the builder pattern, making it easy to construct search parameters.

Here is a an example file:

from __future__ import print_function
from wordplay.dictionary import Dictionary
from wordplay.criteria import Criteria


def main():
    dictionary = Dictionary()

    result = dictionary.get_words_with_any_letters('diction', 6)
    print(result)
    # ['diotic', 'dition', 'indico', 'indict', 'nidiot', 'odinic']

    result = dictionary.get_words_with_any_letters('pox')
    print(result)
    # ['o', 'op', 'ox', 'p', 'po', 'pox', 'x']

    result = dictionary.get_anagrams('aekst')
    print(result)
    # ['keats', 'skate', 'skeat', 'stake', 'steak', 'takes', 'teaks']

    print('car' in dictionary)  # Dictionary is directly iterable
    # True

    criteria = Criteria()
    criteria.begins_with('c').ends_with('s').contains('or')
    criteria.contains_at(('o', 2), ('r', 4)).size_is(10)
    result = dictionary.get_words(criteria)
    print(result)
    # ['corrosives', 'correlates', 'corrugates']


if __name__ == '__main__':
    main()

For further example usage see the Documentation.

To run tests, use pytest.

Documentation

API Reference

License

Apache Software License