按键排序列表

Question

当前正在处理换位问题。我至今是用户输入的消息，该消息被加密成一个列表，像下面：

['BC', 'DE', 'DE', 'DA', 'FD', 'DD', 'BE', 'FE', 'DA', 'EA', 'FE', 'BC'] 

我有密码的下一阶段是把这个变成一个表的关键从用户输入。因此，如果用户输入“CODE”它输出该：

2: Enter the keyword for final encryption: code 
    C O D E 
['B', 'C', 'D', 'E'] 
['D', 'E', 'D', 'A'] 
['F', 'D', 'D', 'D'] 
['B', 'E', 'F', 'E'] 
['D', 'A', 'E', 'A'] 
['F', 'E', 'B', 'C'] 

的下一阶段是取各列的每一个值和打印对应于其列字母的值。所以我的预期输出将是：

C D E O 
['B', 'D', 'E', 'C'] 
['D', 'D', 'A', 'E'] 
['F', 'D', 'D', 'D'] 
['B', 'F', 'E', 'E'] 
['D', 'E', 'A', 'A'] 
['F', 'B', 'C', 'E'] 

我遇到的问题是试图知道如何将每个值放在其相应的列和打印它们。

这是我到目前为止有：

def encodeFinalCipher(): 
    matrix2 = [] 
    # Convert keyword to upper case 
    key = list(keyword.upper()) 

    # Convert firstEncryption to a string 
    firstEncryptionString = ''.join(str(x) for x in firstEncryption) 

    # Print the first table that will show the firstEncryption and the keyword above it 
    keywordList = list(firstEncryptionString) 
    for x in range(0,len(keywordList),len(keyword)): 
     matrix2.append(list(keywordList[x:x+len(keyword)])) 

    # Print the un-ordered matrix to the screen 
    print (' %s' % ' '.join(map(str, key))) 
    for letters in matrix2: 
     print (letters) 

    unOrderedMatrix = [[matrix2[i][j] for i in range(len(matrix2))] for j in range(len(matrix2[0]))] 
    for index, item in enumerate (unOrderedMatrix): 
     print("\n",index, item) 

    index = sorted(key) 
    print(index) 

我得到的排序键的输出：

['A', 'K', 'M', 'R'] 

我想知道的是怎么能这样排序的关键应用到他们代表的价值？我知道我可以通过这样做得到第一列：

print(unOrderedMatrix[0]) 

哪一个得到我第一列的列表。

任何帮助将不胜感激。 Python的

Answer 1

msg = ['BC', 'DE', 'DE', 'DA', 'FD', 'DD', 'BE', 'FE', 'DA', 'EA', 'FE', 'BC', '12'] 
key = 'CODE' 
# 'flatten' the message 
msg = ''.join(msg) 
key_length = len(key) 
#create a dictionary with the letters of the key as the keys 
#use a slice to create the values 
columns = {k:msg[i::key_length] for i, k in enumerate(key)} 
print columns 
# sort the columns on the key letters 
columns = sorted(columns.items()) 
print columns 
# separate the key from the columnar data 
header, data = zip(*columns) 
print header 
# transpose and print 
for thing in zip(*data): 
    print thing 

>>> 
{'C': 'BDFBDF1', 'E': 'EADEAC', 'D': 'DDDFEB', 'O': 'CEDEAE2'} 
[('C', 'BDFBDF1'), ('D', 'DDDFEB'), ('E', 'EADEAC'), ('O', 'CEDEAE2')] 
('C', 'D', 'E', 'O') 
('B', 'D', 'E', 'C') 
('D', 'D', 'A', 'E') 
('F', 'D', 'D', 'D') 
('B', 'F', 'E', 'E') 
('D', 'E', 'A', 'A') 
('F', 'B', 'C', 'E') 
>>> 

Answer 2

这里完整的初学者是什么让你开始（你可能想单独一环衬，以较小位）：

# define data 
data = [['B', 'C', 'D', 'E'], ['D', 'E', 'D', 'A'], ['F', 'D', 'D', 'D'], ['B', 'E', 'F', 'E'], ['D', 'A', 'E', 'A'], ['F', 'E', 'B', 'C']] 

# choose code word 
code = 'code' 

# add original locations to code word [(0, c), (1, o), (2, d), (3, e))] 
# and sort them alphabetically! 
code_with_locations = list(sorted(enumerate(code))) 

print code_with_locations # [(0, 'c'), (2, 'd'), (3, 'e'), (1, 'o')] 

# re-organize data according to new indexing 
for index in range(len(data)): 
    # check if code is shorter than list in current index, 
    # or the other way around, don't exceed either list 
    max_substitutions = min(map(len, [code_with_locations, data[index]])) 

    # create a new list according to new indices 
    new_list = [] 
    for i in range(max_substitutions): 
     current_index = code_with_locations[i][0] 
     new_list.append(data[index][current_index]) 

    # replace old list with new list 
    data[index] = new_list 


print data 

的“代码”输出将是：

[['B', 'D', 'E', 'C'], 
['D', 'D', 'A', 'E'], 
['F', 'D', 'D', 'D'], 
['B', 'F', 'E', 'E'], 
['D', 'E', 'A', 'A'], 
['F', 'B', 'C', 'E']] 

Answer 3

code = raw_input("Enter the keyword for final encryption:") 

user_input = ['BC', 'DE', 'DE', 'DA', 'FD', 'DD', 'BE', 'FE', 'DA', 'EA', 'FE', 'BC'] 
user_input = ''.join(user_input) 

matrix = [user_input[i:i+len(code)] for i in range(0, len(user_input), len(code))] 
matrix.insert(0, code) 

result = sorted([[matrix[j][ind] for j in range(len(matrix))] for ind in range(len(code)) ], key= lambda i:i[0]) 
for row in [[each[ind] for each in result] for ind in range(len(result[0]))]: 
    print row 

打印row结果：

Enter the keyword for final encryption:CODE 
['C', 'D', 'E', 'O'] 
['B', 'D', 'E', 'C'] 
['D', 'D', 'A', 'E'] 
['F', 'D', 'D', 'D'] 
['B', 'F', 'E', 'E'] 
['D', 'E', 'A', 'A'] 
['F', 'B', 'C', 'E'] 

Answer 4

与itertools

from pprint import pprint 
from itertools import chain, izip_longest 
def grouper(iterable, n, fillvalue=None): 
    "Collect data into fixed-length chunks or blocks" 
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx 
    args = [iter(iterable)] * n 
    return izip_longest(fillvalue=fillvalue, *args) 

z = ['BC', 'DE', 'DE', 'DA', 'FD', 'DD', 'BE', 'FE', 'DA', 'EA', 'FE', 'BC'] 
input = 'CODE' 
pprint([[b for (a, b) in sorted(zip(input, x))] 
      for x in grouper(chain.from_iterable(z), len(input))]) 
[['B', 'D', 'E', 'C'], 
['D', 'D', 'A', 'E'], 
['F', 'D', 'D', 'D'], 
['B', 'F', 'E', 'E'], 
['D', 'E', 'A', 'A'], 
['F', 'B', 'C', 'E']]