在Python中用数字替换单词中的多个字母？

Question

我似乎搞不清楚怎么把字母换成数字。比如说，

假设

     'a' , 'b' and 'c' should be replaced by "2".
     'd' , 'e' and 'n' should be replaced by "5".
     'g' , 'h' and 'i' should be replaced by "7".

我想替换的字符串是 again。我想得到的输出是 27275。这些数字的结果应该是字符串形式。

到目前为止，我得到了：

def lett_to_num(word):
    text = str(word)
    abc = "a" or "b" or "c"
    aef = "d" or "e" or "n"
    ghi = "g" or "h" or "i"
    if abc in text:
        print "2"
    elif aef in text:
        print "5"
    elif ghi in text:
        print "7"

^我知道上面的代码是错的^

我应该写什么函数呢？

Answer 1

这样怎么样：

>>> d = {'a': 2, 'c': 2, 'b': 2, 
         'e': 5, 'd': 5, 'g': 7, 
         'i': 7, 'h': 7, 'n': 5}

>>> ''.join(map(str, [d[x] if x in d.keys() else x for x in 'again']))
'27275'
>>>
>>> ''.join(map(str, [d[x] if x in d.keys() else x for x in 'againpp']))
'27275pp'
>>>

Answer 2

这要看情况。因为你似乎是在学习，所以我会避免讲一些高级的库用法。有一种方法可以这样做：

def lett_to_num(word):
    replacements = [('a','2'),('b','2'),('d','5'),('e','5'),('n','5'),('g','7'),('h','7'),('i','7')]
    for (a,b) in replacements:
       word = word.replace(a,b)
    return word

print lett_to_num('again')

还有一种方法，跟你在问题中展示的代码比较接近：

def lett_to_num(word):
    out = ''
    for ch in word:
        if ch=='a' or ch=='b' or ch=='d':
            out = out + '2'
        elif ch=='d' or ch=='e' or ch=='n':
            out = out + '5'
        elif ch=='g' or ch=='h' or ch=='i':
            out = out + '7'
        else:
            out = out + ch
    return out

Answer 3

使用字符串中的maketrans：

from string import maketrans
instr = "abcdenghi"
outstr = "222555777"
trans = maketrans(instr, outstr)
text = "again"
print text.translate(trans)

输出：

 27275

字符串模块中的maketrans可以创建一个字节映射，从输入字符串（instr）到输出字符串（outstr）。当我们使用translate时，如果在输入字符串中找到了某个字符，它就会被替换成输出字符串中对应的字符。