python代码

# 读取.mat文件
mat_data = scipy.io.loadmat('C.mat')
# 获取矩阵数据
adj_matrix = mat_data['C']

一个简短的python代码：

alph=['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
A0=[.082,.015,.028,.043,.127,.022,.020,.061,.070,.0015,.0077,.040,.024,.067,.075,.019,0.001,.060,.063,.091,.028,.0098,.024,.0015,.020,.00074]

#Takes the dot product of two vectors. The inputs V1 and V2 should be list objects.
def DotProduct (V1 : list, V2 : list):
  assert len(V1)==len(V2)
  dp=0
  for i in range(len(V1)):
    dp+=V1[i]*V2[i]
  return dp

# Takes a vector and shifts its elements by a given amount.
#Elements at the end wrap around to the front of the new vector.
def VectorShift( V : list, shift : int):
  S=[]
  for i in range(len(V)):
    index = (i+shift)%len(V)
    S.append(V[index])
  return S

def FrequencyCount( Cipher : str):
  count = [0] * len(alph)
  for letter in Cipher:
    if letter in alph:
      index = alph.index(letter)
      count[index] += 1
  return count

#DotProduct(alph,A0)
a=[1,2,3,4,5]
S=VectorShift(a,2)
print(S)
str='AACCCDZZ';
SS=FrequencyCount(str)
print(SS)

def ShiftCipher(cipher):
    freq = FrequencyCount(cipher)  # 统计字母频率
    max_freq_index = freq.index(max(freq))  # 找到出现频率最高的字母索引
    shift = (max_freq_index - alph.index('E')) % len(alph)  # 计算需要进行的字母位移
    plaintext = ''
    for letter in cipher:
        if letter in alph:
            index = (alph.index(letter) - shift) % len(alph)  # 进行字母位移解密
            plaintext += alph[index]
        else:
            plaintext += letter  # 非字母字符保持不变
    return plaintext

cipher= 'ABCDE'
sss=ShiftCipher(cipher)
print(sss)