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Anagram maker
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__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() |
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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"