I am looking to build a list of words (three letters or more) that make up a large list of six-letter words. I was wondering if there was a cleaner or more efficient way to loop through the characters to build these subsets.

The top of this program (not posted) takes two .txt files - one containing the list of 6-letter words and one containing the Scrabble Dictionary (used to determine what is a real word).

The main core of my code is below. I loop through each character of the word, making sure indexes don't repeat and then build each word subset. I then sort the list on length and convert it into a string via .join to feed into another app I am writing for "subset guessing".

# Define Dictionary Lookup Function
def isDictionaryWord(word):
    if (word.lower().strip()) in listDictionary:
        return True
        return False

def createSubsets(word):
    print 'Starting Word Subset - ' + word
    starttime = datetime.datetime.now()
    tempOutput = []

    ## 1-Letter Word Loop
    for c in range(0, len(word)):

        ## 2-Letter Word Loop (skip)
        for d in range(0, len(word)):
            if (c != d):

                    ## 3-Letter Word Loop
                    for e in range(0, len(word)):
                        if (c != e) and (d != e):
                            threeLetterWord = word[c] + word[d] + word[e]
                            if isDictionaryWord(threeLetterWord) and (threeLetterWord not in tempOutput):
                                #print threeLetterWord

                            ## 4-Letter Word Loop
                            for f in range(0, len(word)):
                                if (c != f) and (d != f) and (e != f):
                                    fourLetterWord = word[c] + word[d] + word[e] + word[f]
                                    if isDictionaryWord(fourLetterWord) and (fourLetterWord not in tempOutput):
                                        #print fourLetterWord

                                    ## 5-Letter Word Loop
                                    for g in range(0, len(word)):
                                        if (c != g) and (d != g) and (e != g) and (f != g):
                                            fiveLetterWord = word[c] + word[d] + word[e] + word[f] + word[g]
                                            if isDictionaryWord(fiveLetterWord) and (fiveLetterWord not in tempOutput):
                                                #print fiveLetterWord

    endtime = datetime.datetime.now()

    ## Sort Array by Length for Game Purposes & Convert to String
    tempOutput.sort(key = len)
    tempOutputStr = word + ": " + ", ".join(tempOutput)

    print 'Total Time Taken (s) - ' + str((endtime - starttime).total_seconds())
    print 'Total Subset Words   - ' + str(len(tempOutput) - 1) + '\n'
    print 'Temp Output Array    - ' + str(tempOutput) + '\n'
    print 'Temp Output String   - ' + tempOutputStr
    print '========================================'

    # Print tempOutputStr
    fOut.write(tempOutputStr + '\n')

# Loop through all 6-letter words and create subsets
for word in listSix:

1 Answer 1


Here are some comments, in no particular order:

  • You can simplify the definition of isDictionaryWord():

    def isDictionaryWord(word):
        return word.lower().strip() in listDictionary

    although as it stands, that won’t work because listDictionary isn’t defined.

    Speaking of that function, this comment:

    # Define Dictionary Lookup Function

    tells me nothing I couldn’t learn from reading the next line. It’s superfluous and should be removed.

  • Have a read of PEP 8, particularly with regards to variable naming. Python uses snake_case for most variables, with CamelCase reserved for classes only.

  • Since you’re using Python 2.x (I can tell from the print statements), it’s more efficient to use xrange() instead of range() – this creates a memory-efficient iterator rather than creating a list, which can make your code much faster over large ranges.

  • You could tidy up some of your checks by making tempOutput a set instead of a list, then uniqueness is enforced by Python. Your innermost checks become more like:

    if is_dictionary_word(three_letter_word):

  • Rather than writing out the loops explicitly, I’d suggest looking at the itertools module. This contains some very fast functions for doing big loops. For example, your check for three-letter words reduces to the following:

    import itertools
    temp_output = []
    for tmp_word in itertools.permutations(word.lower(), r=3):
        if is_dictionary_word(tmp_word):

    You can add similarly simple loops to get the four and five-letter words.

    This has a few other benefits:

    • This approach enforces uniqueness, so we can use a list (which preserves ordering) without needing to check if a word is already there.
    • The array is already sorted by length of the word, with the shortest words first.
    • By lowercasing the main word once, we can speed up our calls to is_dictionary_word by getting rid of the .lower() and .strip() calls.

  • There’s too much code in the createSubsets() function. That function includes:

    • Printing debug code about what it’s doing, and how long it took
    • Finding a list of subset words
    • Writing that list to a file

    It would be better if each of those tasks were handled by separate functions – this would allow you to reuse code more easily among different parts of your code.


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