我正在尝试创建一个凯撒密码的频率分析，但我似乎无法得到输出

Question

我正在写一个代码，我想计算密文中使用字母的频率，然后将其与ETAOINSHRDLCUMWFGYPBVKJXQZ进行比较。然后我想让它建议一个密钥使用解密时，但我不能从我目前的编码输出。你知道吗

englishLetterFreq = {'E', 'T', 'A', 'O', 'I', 'N', 'S', 'H', 'R', 'D', 'L', 'C', 'U', 'M', 'W', 'F', 'G', 'Y', 'P', 'B', 'V', 'K', 'J', 'X', 'Q', 'Z'}
ETAOIN = 'ETAOINSHRDLCUMWFGYPBVKJXQZ'
LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'


def getLetterCount(message):
    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
            print('LETTERS {}'.format(letterCount))
    return letterCount

def getItemAtIndexZero(x):
    return x[0]


def getFrequencyOrder(message):
    letterToFreq = getLetterCount(message)
    letterToFreq = {}
    for letter in LETTERS:
        if letterToFreq[letter] not in freqToletter:
            freqToLetter[letterToFreq[letter]] = [letter]
        else:
            freqToLetter[letterToFreq[letter]].append(letter)

    for freq in freqToLetter:
        freqToLetter[freq].sort(key=ETAOIN.find, reverse=True)
        freqToLetter[freq] = ".join(freqToLetter[freq])"
        freqPairs = list(freqToLetter.items())
        freqPairs.sort(key=getItemAtIndexZero, reverse=True)
        freqOrder = []
        for freqPair in freqPairs:
            freqOrder.append(freqPair[1])
        return".join(freqOrder)"


def englishFreqMatchScore(message):

    freqOrder = getFrequencyOrder(message)
    matchScore = 0
    for commonLetter in ETAOIN[:6]:
        if commonLetter in freqOrder[:6]:
            matchScore += 1
    for uncommonLetter in ETAOIN[-6:]:
        if uncommonLetter in freqOrder[-6:]:
            matchScore += 1
            print("{}",englsishFreqMatchScore)
    return matchScore

Answer 1

请看以下内容。我很抱歉没有纠正你的代码，但有时重写一些东西比纠正它容易。有什么不清楚的尽管问我。你知道吗

from collections import Counter


def decr(text, jump):
    alpha = 'abcdefghijklmnopqrstuvwxyz'
    decryption = ''
    for char in text.lower():
        ind = alpha.find(char)
        if ind != -1:
            decryption += alpha[ind - jump]
        else:
            decryption += char
    return decryption

to_dec = r'Max yheehpbgz mhhe teehpl rhn mh xgvkrim t mxqm pbma t lbfiex hyylxm tezhkbmaf - telh dghpg tl Vtxltk vbiaxk. By rhn tkx nlbgz 13 tl max dxr, max kxlnem bl lbfbetk mh tg khm13 xgvkrimbhg. By rhn nlx "znxll" tl max dxr, max tezhkbmaf mkbxl mh ybgw max kbzam dxr tgw wxvkriml max lmkbgz ur znxllbgz. B telh pkhmx t lftee tkmbvex (pbma lhnkvx inuebvtmbhg) tuhnm ybgwbgz max kbzam dxr bg tg ngdghpg vhgmxqm hy tg xgvkrimxw mxqm. By rhn ptgm mh dghp fhkx, B abzaer kxvhffxgm mabl uhhd.'

alpha = 'abcdefghijklmnopqrstuvwxyz'
freq = 'etaoinsrhdlucmfywgpbvkxqjz'
n = 3

c = Counter(''.join(x for x in to_dec.lower() if x in alpha))
order = ''.join(x[0] for x in c.most_common())

f = lambda x: alpha.find(x[0]) - alpha.find(x[1])
nums = [f(x) if f(x) >= 0 else f(x) + 26 for x in zip(order, freq)]

d = Counter(nums)

for i in d.most_common(n):
    key = i[0]
    print('Trying Ceasar with n = {}'.format(key))
    print(decr(to_dec, key))
    print()

令人失望的是（但并不奇怪），每次运行此代码时，可能会得到不同的结果。这是不允许的，但发生这种情况的原因是collections.Counter().most_common()当to项具有相同的计数时，它以任意顺序返回它们。但是，随着要加密的文本大小的增加，两个字母的计数完全相同的可能性也就消失了。将n设置为1也有帮助（代码为ceaar键提供了一个建议）。你知道吗