freqAnalysis.py

englishLetterFreq = {'E': 12.70, 'T': 9.06, 'A': 8.17, 'O': 7.51, 'I': 6.97, 'N': 6.75, 'S': 6.33, 'H': 6.09, 'R': 5.99, 'D': 4.25, 'L': 4.03, 'C': 2.78, 'U': 2.76, 'M': 2.41, 'W': 2.36, 'F': 2.23, 'G': 2.02, 'Y': 1.97, 'P': 1.93, 'B': 1.29, 'V': 0.98, 'K': 0.77, 'J': 0.15, 'X': 0.15, 'Q': 0.10, 'Z': 0.07}
ETAOIN = 'ETAOINSHRDLCUMWFGYPBVKJXQZ'
LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'



def getLetterCount(message):
    # Returns a dictionary with keys of single letters and values of the
    # count of how many times they appear in the message parameter.
    letterCount = {'A': 0, 'B': 0, 'C': 0, 'D': 0, 'E': 0, 'F': 0, 'G': 0, 'H': 0, 'I': 0, 'J': 0, 'K': 0, 'L': 0, 'M': 0, 'N': 0, 'O': 0, 'P': 0, 'Q': 0, 'R': 0, 'S': 0, 'T': 0, 'U': 0, 'V': 0, 'W': 0, 'X': 0, 'Y': 0, 'Z': 0}

    for letter in message.upper():
        if letter in LETTERS:
            letterCount[letter] += 1

    return letterCount

def getItemAtIndexZero(x):
    return x[0]

def getFrequencyOrder(message):

    # Dictionary of each letter and its frequency count
    letterToFreq = getLetterCount(message)

    # Dictionary of each frequency count to each letter(s) with that frequency
    freqToLetter = {}
    for letter in LETTERS:
        if letterToFreq[letter] not in freqToLetter:
            freqToLetter[letterToFreq[letter]] = [letter]
        else:
            freqToLetter[letterToFreq[letter]].append(letter)

    # Put each list of letters in reverse "ETAOIN" order, and then convert it to a string
    for freq in freqToLetter:
        freqToLetter[freq].sort(key=ETAOIN.find,reverse=True)
        freqToLetter[freq] = ''.join(freqToLetter[freq])

    # Convert the freqToLetter dictionary to a list of tuple pairs (key, value), then sort them
    freqPairs = list(freqToLetter.items())
    freqPairs.sort(keys=getItemAtIndexZero,reverse=True)

    # Now that the letters are ordered by frequency, extract all the letters for the final string
    freqOrder = []
    for freqPair in freqPairs:
        freqOrder.append(freqPair[1])

    return ''.join(freqOrder)

def englishFreqMatchScore(message):
    # Return the number of matches that the string in the message
    # parameter has when its letter frequency is compared to English
    # letter frequency. A "match" is how many of its six most frequent
    # and six least frequent letters is among the six most frequent and
    # six least frequent letters for English.
    freqOrder = getFrequencyOrder(message)

    matchScore = 0
    # Find how many matches from six most common letters
    for commonLetter in ETAOIN[:6]:
        if commonLetter in freqOrder[:6]:
            matchScore += 1
    # Find how many matches for the size least common letters
    for uncommonLetter in ETAOIN[-6:]:
        if uncommonLetter in freqOrder[-6:]:
            matchScore += 1

    return matchScore