This is my code for anagrams not utilizing the ready prepared file of anagram synonyms to celebrate 11.11.11 11:11:11.
If you start program it warns you about missing subdirectory dict and creates it. You may put any word files, one word per line, to file <dictionary name>.txt. Program lists available directories if started in interactive prompt mode by not supplying parameters from command line. Few usefull dictionary locations mentioned also in code:
""" Major learning project of Tony 'pyTony' Veijalainen 2010-2011 keep attribution of source if you use this code elsewhere. The most clear code version without special tricks or anagram word synonyms dictionary. Published 11.11.11 at DaniWeb Python Forum Usage: python sANAsTony.py uk 1 DaniWeb Forum # answer prompts interactively python sANAsTony.py """ import sys import os if sys.version < '3': input, range = raw_input, xrange try: import psyco; psyco.full() except: print('(Psyco would improve performance noticably in Python 2.6 and earlier)') def contains(bigger,smaller): """ find the letters that are left from bigger, when smaller's letters are taken from bigger or return None, if smaller's letters are not contained in bigger """ if len(bigger) >= len(smaller): while smaller: this, smaller = smaller[0:1] , smaller[1:] if this not in bigger: return None bigger = bigger.replace(this, '', 1) return bigger takeout=" \t'-+\n\r" def trim_word(word, takeout=takeout): """ lowercase the word and return cleaned word """ #word with letters not in takeout return word.lower().translate(None, takeout) # if you have clean directory and do not mind "we'd" missing use next line instead. # return word.lower().strip() def find_words(candidate_words, letters, atleast): """ candidate_words is iterable giving the words to choose from, like open file or list of words """ valid_words =  for this in candidate_words: # we do not assume clean or ordered words, this is costly in execution time, # but more flexible this = trim_word(this) if contains(letters, this) is not None: if len(letters) >= len(this) >= atleast: valid_words.append(this) return sorted(valid_words, key=len, reverse=True) def find_anagrams(word, words, atleast): """ Find possibly multiple word anagrams from parameter 'word' with possibly repeating words of 'atleast' or more letters from sequence parameter 'words' in order of the sequence (combinations with repeats, not permutations) """ for word_index, current_word in enumerate(words): remaining_letters = contains(word, current_word) if remaining_letters=='': yield current_word elif remaining_letters is not None and len(remaining_letters) >= atleast: for x in find_anagrams(remaining_letters, words[word_index:], atleast): yield (' '.join((current_word, x))) if __name__ == '__main__': if not os.path.isdir('dict'): os.mkdir('dict') raise SystemExit(''' No dictionaries installed, copy wordlists like wordlist from http://wordlist.sourceforge.net/ or scrable wordslist http://www.isc.ro/en/commands/lists.html to dict subdirectory!''') if len(sys.argv) > 3: print('Taking language, smallest word and the words (rest of line) from command line.') language, smallest, words = (sys.argv, int(sys.argv), trim_word(''.join(sys.argv[3:]))) else: words = trim_word(input('Give words: ')) smallest = int(input('Minimum acceptable word length: ')) print('\n\t' + '\n\t'.join(d[:-4] for d in os.listdir('dict') if d.endswith('.txt'))) language = input("Language ('.txt' added automatically): ") language_file = 'dict/%s.txt' % language if not os.path.isfile(language_file): print('%s does not exist in dict directory.' % language) else: with open(language_file) as word_file: wordlist = find_words(word_file, words, smallest) print('%i words loaded.' % len(wordlist)) # print out solution solution_number = 0 for solution_number, word in enumerate(find_anagrams(words, wordlist, smallest), 1): print("%5i: %s" % (solution_number, word)) print('\n%i solutions found!' % solution_number)