'leet'程序-获取所有排列

问题描述

我写了一个简单的函数，可以将单词中的字母转换为"leet"数字对应物.

I wrote a straightforward function that converts letters in words to their 'leet' numerical counterparts.

def Leet(word):
    letters = list(word.lower())
    for n, letter, in enumerate(letters):
        if letter == 'o':
            letters[n]= '0'
        elif letter == 'i':
            letters[n]= '1'
        elif letter == 'z':
            letters[n]= '2'
        elif letter == 'e':
            letters[n]= '3'
        elif letter == 'a':
            letters[n]= '4'
        elif letter == 's':
            letters[n]= '5'
        elif letter == 'g':
            letters[n]= '6'
        elif letter == 't':
            letters[n]= '7'
        elif letter == 'b':
            letters[n]= '8'
    return ''.join(letters)

因此，当我输入'zit'时，程序将返回'217'.

so when I input 'zit', the program will return '217'.

我的问题是，如何更改它以提供所有可能的排列方式('217'，'2it'，'z1t'，'zi7'，'21t'等)?我已经读过有关 itertools 的信息，但对于如何将其应用到我的函数中却感到困惑.

My question is, how can I change it to give me every possible permutation ('217', '2it', 'z1t', 'zi7', '21t', etc.)? I've read about itertools but I am stumped as to how to apply it to my function.

推荐答案

第一个观察结果是您可以缩短查找，如下所示:

First observation is that you can shorten the lookup, like this:

REPLACE = { letter: str(index) for index, letter in enumerate('oizeasgtb') }

def Leet2(word):
    letters = [ REPLACE.get(l, l) for l in word.lower() ]
    return ''.join(letters)

REPLACE 如下:

{'a': '4', 'b': '8', 'e': '3', 'g': '6', 'i': '1',
 'o': '0', 's': '5', 't': '7', 'z': '2'}

REPLACE.get(l，l)会给您发回替换字母，或者如果没有替换字母，则返回原始字母.

And the REPLACE.get(l,l) gives you back either the replacement letter, or the original letter if there is no replacement.

第二个观察结果是您并不真正想要排列，排列是在移动.'217'的排列是:

The second observation is that you don't really want permutations, which are shifts in ordering. The permutations of '217' are:

>>> [ ''.join(p) for p in permutations('217') ]
['217', '271', '127', '172', '721', '712']

您真正需要的是一个列表的产品，该列表对给定字符位置的所有可能选择进行编码:

What you really need is the product of a list that encodes all the possible choices for a given character position:

[('z', '2'), ('i', '1'), ('t', '7')]

如果我还显示带有某些字符的可能性列表，而这些字符没有有效替代，则可能会更清楚地了解其工作原理.例如，对于'red':

How this works might be clearer if I also show a possibility list with some characters for which there is no valid replacement. For 'red' for example:

[('r',), ('e', '3'), ('d',)]

现在，我们需要这些选项的字符串连接产品.全部放在一起:

Now we need the string-joined product of those options. Putting it all together:

from itertools import product

def Leet2Combos(word):
    possibles = []
    for l in word.lower():
        ll = REPLACE.get(l, l)
        possibles.append( (l,) if ll == l else (l, ll) )
    return [ ''.join(t) for t in product(*possibles) ]

print Leet2Combos('zit')
print Leet2Combos('red')

赠予:

['zit', 'zi7', 'z1t', 'z17', '2it', '2i7', '21t', '217']
['red', 'r3d']

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