将两个相似的函数泛化为一个答案

【问题标题】：Generalizing two similar functions into one将两个相似的函数泛化为一个
【发布时间】：2015-01-07 04:11:23
【问题描述】：

所以下面是一些能够编码和解码句子的 Python 代码。我不确定如何定义一个函数，该函数采用三个参数，可以概括或组合这两个函数：mapInToOutEncode 和 mapInToOutDecode 合二为一。

alphabet = " abcdefghijklmnopqrstuvwxyz"
keyString ="kjihgfedcba zyxwvutsrqponml"

def mapInToOutEncode(inChar):
    index = 0
    found = 0
    for char in alphabet:
        if(char == inChar):
            found = index #remember it in a variable found
        else:
            index +=1
    outChar = keyString[found]
    return(outChar)

def mapInToOutDecode(inChar):
    index = 0
    found = 0
    for char in alphabet:
        if(char == inChar):
            found = index #remember it in a variable found
        else:
            index +=1
    outChar = keyString[found]
    return(outChar)

def encode(inString):
    outString = ''
    for char in inString:
        outString += mapInToOutEncode(char)
    return(outString)

def decode(inString):
    outString = ''
    for char in inString:
        outString += mapInToOutDecode(char)
    return(outString)

print( encode("this is a crazy bit of news to share we attack at dawn" ))
print( decode(encode("this is a crazy bit of news to share we attack at dawn")))

【问题讨论】：

这实际上不是“如何组合功能”的答案，但您正在寻找的功能已经在str.translate() 中实现，另请参见string.maketrans()。

标签： python string python-3.x decode encode

【解决方案1】：

首先，请注意您的函数mapInToOutEncode 和mapInToOutDecode 是相同的。为了获得您想要的功能，mapInToOutDecode 应该是：

def mapInToOutDecode(inChar):
    index = 0
    found = 0
    for char in keyString:
        if(char == inChar):
            found = index #remember it in a variable found
            break
        else:
            index +=1
    outChar = alphabet[found]
    return(outChar)

请注意，我还添加了单词break，因此您的代码一旦找到它就不会继续寻找inchar。

现在，要获得您想要的（mapInToOutEncode 和 mapInToOutDecode 的通用函数），您可以：

def mapInToOut(inChar, alphFrom, alphTo):
    index = 0
    found = 0
    for char in alphFrom:
        if(char == inChar):
            found = index #remember it in a variable found
            break
        else:
            index +=1
    outChar = alphTo[found]
    return(outChar)

要编码每个字符a，您可以使用mapInToOut(a, alphabet, keyString)。要解码每个字符b，您可以使用mapInToOut(b, keyString, alphabet)。

所以，encode 和 decode 函数看起来像这样：

def encode(inString):
    outString = ''
    for char in inString:
        outString += mapInToOut(char, alphabet, keyString)
    return(outString)

def decode(inString):
    outString = ''
    for char in inString:
        outString += mapInToOut(char, keyString, alphabet)
    return(outString)

这是一个完整工作示例（即，这应该是您的全部代码）：

alphabet = " abcdefghijklmnopqrstuvwxyz"
keyString ="kjihgfedcba zyxwvutsrqponml"

def mapInToOut(inChar, alphFrom, alphTo):
    index = 0
    found = 0
    for char in alphFrom:
        if(char == inChar):
            found = index #remember it in a variable found
            break
        else:
            index +=1
    outChar = alphTo[found]
    return(outChar)

def encode(inString):
    outString = ''
    for char in inString:
        outString += mapInToOut(char, alphabet, keyString)
    return(outString)

def decode(inString):
    outString = ''
    for char in inString:
        outString += mapInToOut(char, keyString, alphabet)
    return(outString)

print ("this is a crazy bit of news to share we attack at dawn")
encoded_msg = encode("this is a crazy bit of news to share we attack at dawn")
print encoded_msg
print decode(encoded_msg)

【讨论】：

感谢您指出！当我将 "outChar = keyString[found]" 中的 keyString 更改为字母表时，我从编码和解码得到的输出现在完全一样！而且当我使用您的更改运行它时，它会显示“NameError：未定义全局名称'alpFrom'”......？
那是我 print( encode("这是一个疯狂的新闻分享我们在黎明攻击" )) 和 print( decode(encode("这是一个疯狂的新闻分享我们在黎明攻击")))，我得到 rcbskbskjkhtjlmkibrkwekxfoskrwkscjtfkofkjrrjh kjrkgjox 用于两个打印语句....
我没有在任何地方写过alpFrom。你是用 `ctrl+c , ctrl+v' 复制我的代码还是重新输入？我想你可能抄错了。确保您的代码与我的完全一样（这也适用于您的其他问题）。
它不工作......所以我有 mapInToOut 部分，我还保留了 def encode(inString) 和 def decode(inString) 部分，我是否将我的 for 循环更改为“outString + = mapInToOut(char,alphabet,keyString)" ?
我编辑了我的代码以更好地阐明它。它工作正常，我刚刚测试过。

【解决方案2】：

尝试使用字典进行编码/解码，它将大大简化您的代码：

>>> alphabet = " abcdefghijklmnopqrstuvwxyz"
>>> keyString ="kjihgfedcba zyxwvutsrqponml"
>>> encoder = dict(zip(alphabet, keyString))
>>> decoder = dict(zip(keyString, alphabet))
>>> a = 'this is a string'
>>> q = ''.join(encoder.get(i, i) for i in a)
>>> q
'rcbskbskjksrtbxd'
>>> ''.join(decoder.get(i, i) for i in q)
'this is a string'

【讨论】：