Question

这是表格http://en.wikipedia.org/wiki/File:Vigen%C3%A8re_square_shading.svg

如何在Python中实现此表？这是什么方式？

关于如何检查例如Row：L＆amp; G栏成为R欢迎。

Answer 1

>>> letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
>>> mat = [(letters * 2)[i:i+26] for i in range(26)]
>>> print '\n'.join(mat)
ABCDEFGHIJKLMNOPQRSTUVWXYZ
BCDEFGHIJKLMNOPQRSTUVWXYZA
CDEFGHIJKLMNOPQRSTUVWXYZAB
DEFGHIJKLMNOPQRSTUVWXYZABC
EFGHIJKLMNOPQRSTUVWXYZABCD
FGHIJKLMNOPQRSTUVWXYZABCDE
GHIJKLMNOPQRSTUVWXYZABCDEF
HIJKLMNOPQRSTUVWXYZABCDEFG
IJKLMNOPQRSTUVWXYZABCDEFGH
JKLMNOPQRSTUVWXYZABCDEFGHI
KLMNOPQRSTUVWXYZABCDEFGHIJ
LMNOPQRSTUVWXYZABCDEFGHIJK
MNOPQRSTUVWXYZABCDEFGHIJKL
NOPQRSTUVWXYZABCDEFGHIJKLM
OPQRSTUVWXYZABCDEFGHIJKLMN
PQRSTUVWXYZABCDEFGHIJKLMNO
QRSTUVWXYZABCDEFGHIJKLMNOP
RSTUVWXYZABCDEFGHIJKLMNOPQ
STUVWXYZABCDEFGHIJKLMNOPQR
TUVWXYZABCDEFGHIJKLMNOPQRS
UVWXYZABCDEFGHIJKLMNOPQRST
VWXYZABCDEFGHIJKLMNOPQRSTU
WXYZABCDEFGHIJKLMNOPQRSTUV
XYZABCDEFGHIJKLMNOPQRSTUVW
YZABCDEFGHIJKLMNOPQRSTUVWX
ZABCDEFGHIJKLMNOPQRSTUVWXY

或者，您可以动态计算元素：

def get_elem(row, col):
  return 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[(ord(row) + ord(col)) % 26]

print get_elem('L', 'G')

Answer 2

我喜欢在可能的情况下使用标准库，以避免滚动我自己的代码并无意中遗漏（比如意外错位或忘记了一封信。）

import string
import collections

def vigsquare(printable=False):
    '''
    Returns a string like a vigenere square,
    printable joins each row with a newline so it's literally square
    printable=False (defaul) joins without newlines for easier 
    searching by row and column index
    '''
    alpha = string.ascii_uppercase
    rotater = collections.deque(alpha)
    vigsquare_list = []
    for i in xrange(26):
        vigsquare_list.append(''.join(rotater))
        rotater.rotate(-1)
    if printable:
        return '\n'.join(vigsquare_list) 
    else:
        return ''.join(vigsquare_list)

和

def vigenere(row, column):
    ''' 
    Return a character from a vigenere square by 
    row and column letter.
    vigenere('L', 'G') returns 'R'
    '''
    alpha = string.ascii_uppercase
    rowindex = alpha.find(row)
    columnindex = alpha.find(column)
    return vigsquare()[rowindex*26 + columnindex]

print vigsquare(printable=True)
vigenere('L', 'G')

给你：

ABCDEFGHIJKLMNOPQRSTUVWXYZ
BCDEFGHIJKLMNOPQRSTUVWXYZA
CDEFGHIJKLMNOPQRSTUVWXYZAB
DEFGHIJKLMNOPQRSTUVWXYZABC
EFGHIJKLMNOPQRSTUVWXYZABCD
FGHIJKLMNOPQRSTUVWXYZABCDE
GHIJKLMNOPQRSTUVWXYZABCDEF
HIJKLMNOPQRSTUVWXYZABCDEFG
IJKLMNOPQRSTUVWXYZABCDEFGH
JKLMNOPQRSTUVWXYZABCDEFGHI
KLMNOPQRSTUVWXYZABCDEFGHIJ
LMNOPQRSTUVWXYZABCDEFGHIJK
MNOPQRSTUVWXYZABCDEFGHIJKL
NOPQRSTUVWXYZABCDEFGHIJKLM
OPQRSTUVWXYZABCDEFGHIJKLMN
PQRSTUVWXYZABCDEFGHIJKLMNO
QRSTUVWXYZABCDEFGHIJKLMNOP
RSTUVWXYZABCDEFGHIJKLMNOPQ
STUVWXYZABCDEFGHIJKLMNOPQR
TUVWXYZABCDEFGHIJKLMNOPQRS
UVWXYZABCDEFGHIJKLMNOPQRST
VWXYZABCDEFGHIJKLMNOPQRSTU
WXYZABCDEFGHIJKLMNOPQRSTUV
XYZABCDEFGHIJKLMNOPQRSTUVW
YZABCDEFGHIJKLMNOPQRSTUVWX
ZABCDEFGHIJKLMNOPQRSTUVWXY

和

'R'

这里有另一个看起来很有趣的解决方案，但我不相信它，因为我不理解它，所以我会对我的代码和其他代码进行单元测试，看看它是否可靠。

def vig_2(row, col):
    return string.ascii_uppercase[(ord(row) + ord(col)) % 26]

比较两种方法的单位测试。第二个（从NPE借来，我现在欠他一个upvote）如果它是正确的话可能会更有效率。：

import unittest
class VigTestCase(unittest.TestCase):
    def test_vigenere(self):
        self.assertEqual(vigenere('L', 'G'), 'R')
    def test_vigsquare(self):
        self.assertEqual(vigsquare(printable=False), 'ABCDEFGHIJKLMNOPQRSTUVWXYZBCDEFGHIJKLMNOPQRSTUVWXYZACDEFGHIJKLMNOPQRSTUVWXYZABDEFGHIJKLMNOPQRSTUVWXYZABCEFGHIJKLMNOPQRSTUVWXYZABCDFGHIJKLMNOPQRSTUVWXYZABCDEGHIJKLMNOPQRSTUVWXYZABCDEFHIJKLMNOPQRSTUVWXYZABCDEFGIJKLMNOPQRSTUVWXYZABCDEFGHJKLMNOPQRSTUVWXYZABCDEFGHIKLMNOPQRSTUVWXYZABCDEFGHIJLMNOPQRSTUVWXYZABCDEFGHIJKMNOPQRSTUVWXYZABCDEFGHIJKLNOPQRSTUVWXYZABCDEFGHIJKLMOPQRSTUVWXYZABCDEFGHIJKLMNPQRSTUVWXYZABCDEFGHIJKLMNOQRSTUVWXYZABCDEFGHIJKLMNOPRSTUVWXYZABCDEFGHIJKLMNOPQSTUVWXYZABCDEFGHIJKLMNOPQRTUVWXYZABCDEFGHIJKLMNOPQRSUVWXYZABCDEFGHIJKLMNOPQRSTVWXYZABCDEFGHIJKLMNOPQRSTUWXYZABCDEFGHIJKLMNOPQRSTUVXYZABCDEFGHIJKLMNOPQRSTUVWYZABCDEFGHIJKLMNOPQRSTUVWXZABCDEFGHIJKLMNOPQRSTUVWXY')
    def test_vig2(self):
        for i in string.ascii_uppercase:
            for j in string.ascii_uppercase:
                self.assertEqual(vig_2(i, j), vigenere(i, j))

unittest.main()
...
----------------------------------------------------------------------
Ran 3 tests in 0.038s

OK

所以看起来NPE有一个非常好的解决方案！

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