Anagram maker

Anagram maker

__author__ = 'danlmarmot'

from itertools import permutations
import collections

INPUT_WORD = "deposit"
MIN_ANAGRAM_LEN = 4
WORD_DICT_FILEPATH = '/usr/share/dict/words'


def main():
    words_as_dict = make_word_dict(WORD_DICT_FILEPATH)
    look_for_anagrams(INPUT_WORD, words_as_dict, min_length=MIN_ANAGRAM_LEN)


def make_word_dict(word_filepath):
    word_dict = collections.defaultdict(lambda: 0)

    for w in open(word_filepath):
        #strip off the /n character with w.strip()
        word_dict[w.lower().strip()] += 1

    return word_dict


def look_for_anagrams(input_word, word_dict, min_length=4):
    anagram_count = 0

    report_header = "Anagrams for " + input_word + ":"
    print report_header
    r = open("results.txt", "w")
    r.write(report_header + "\n")

    for i in range(min_length, len(input_word) + 1):
        # reset this for each string length
        answers_for_strlen = ([])

        for item in permutations(input_word, i):
            test_word = ''.join(item)

            if test_word in word_dict.keys():
                if test_word not in answers_for_strlen and test_word != input_word:
                    anagram_count += 1
                    print test_word
                    answers_for_strlen.append(test_word)
                    r.write(test_word + "\n")

    report_summary = "Total number of anagrams found for '" + input_word + "': " + str(anagram_count)
    print report_summary
    r.write(report_summary + "\n")
    r.close()

if __name__ == '__main__':
    main()

This makes anagrams based on a provided word. Anagrams have to be at least 4 characters in length.

Tested against Python 2.7.6 on MacOS X, with the word dictionary at /usr/share/dict/words and the initial word of "deposit"