Python：使用关键字和字母表的其余部分填充5x5字符矩阵

Question

Programming the Playfair cipher in case anyone has heard about it.

I ask the user to input a keyword (Example: bookkeeping), I then remove repeated letters (Example: bokeping). I need to create a 5x5 matrix and fill it with chars but declaring

    matrix = [][]

is marked as "Invalid syntax". I then need to fill up the matrix with the unique characters and then the rest of the letters of the alphabet (after removing the unique keyword characters) to end up with (i and j are placed on the same position):

    [b, o, k, e, p]
    [ij, n, g, a, c]
    [d, f, h, l, m]
    [q, r, s, t, u]
    [v, w, x, y, z]

Question 1: How do I properly declare the matrix to use it for this?

Question 2: How do I fill up the matrix with the keyword and then the rest of the characters?

Other than that, any suggestions?

Thank you for your help.

Answer 1

Python has groupby :

>>> from itertools import groupby
>>> s = 'aabbccc'
>>> [k for k, g in groupby(s)]
['a', 'b', 'c']

Question 1: How do I properly declare the matrix to use it for this?

A matrix is a list of lists. You declare it like this: m = [[x,y] for x in range(3) for y in range(3)]

Answer 2

You can create a matrix with a nested list structure.

For example the following code defines a 3x3 matrix.

M = [[1, 2, 3],
     [4, 5, 6],
     [7, 8, 9]]

M[1]  # Row 2

[4, 5, 6]

M[1][2]  # Row 2, item 3

6

You can use a list comprehension to process the structure.

[row[1] for row in M]    # Column 2

[2, 5, 8]

Answer 3

For the purpose of the exercise you should just replace all j characters with i in both the pass phrase and cleartext, then you dont need to worry about the double character.

Then to build the matrix you can do the following:

from itertools import chain

key = "bookkeeping".replace("j", "i")
alphabet = "abcdefghiklmnopqrstuvwxyz" # note no J

# dedup the list of characters
slots = []
for x in chain(key, alphabet):
  if not x in slots: slots.append(x)

# get our matrix using list slices
matrix = [ slots[i:i+5] for i in xrange(0, 25, 5) ]
print matrix

Result:

[['b', 'o', 'k', 'e', 'p'], ['i', 'n', 'g', 'a', 'c'], ['d', 'f', 'h', 'l', 'm'], ['q', 'r', 's', 't', 'u'], ['v', 'w', 'x', 'y', 'z']]

Hadn't come across playfair before - is a nice little algorithmic challenge!

Answer 4

If you are paranoid about libraries like I am, then this is the solution of you.

def raw_inp(msg):
    #This function makes sure the raw_input function is available in
    #both python 2.x or 3.x
    try:
        return raw_input(msg)
    except NameError:
        return input(msg)

def unique_array(arr):
    #This function removes duplicates of array values
    check = []
    for i in arr:
        if i not in check:
            check.append(i)
    return check

def form(key):
    key = key.lower() #make sure key is all lowercase
    alpha = []
    key_arr = [] #Convert key to array
    key_arr.extend(key)#Convert key to array
    key = key_arr #Convert key to array
    alpha.extend("abcdefghijklmnopqrstuvwxyz") #Convert alphabet to array
    arr = (key+alpha)
    new_arr = []
    finish_arr = []
    return_arr = []
    arr = unique_array(arr) #Removes any duplicate letters before we start
    for i in arr:
        if i in alpha: #if the character is in the alphabet
            if i == "i" or i == "j": #if it is i or j, place it together
                new_arr.append("ij")
            else:
                new_arr.append(i) #or else place the letter in the array
    x = 0 #array index
    i = 0
    new_arr = unique_array(new_arr)
    while x < 5:
        finish_arr.append([])
        finish_arr[x] = []
        y = 0
        while y < 5:
            finish_arr[x].append(new_arr[i])
            y += 1
            i += 1
        x += 1
    for i in finish_arr:
        return_arr.append(unique_array(i))
    return return_arr


key = raw_inp("Key: ") #get key

arr = (form(key)) #

for i in arr:
    print i

Sorry about the weird style. I programmed it in another way then switched to this way.

This code was meant to be understandable, not short or efficient. Please comment if you don't understand something