打印键，如果满足条件则打印键值

Question

我想创建简单的“编码脚本”。

我有这本词典：

diction =  {

"A" :  "Z", 

"Y" :  "B",

"C" :  "X"

}

我想给出一些随机句子，迭代它的字母，如果在这本字典中找到字母 - 打印相反的字母

所以，如果我输入单词

"ABC"

它应该返回：

"ZYX"

我尝试了这段代码，但我有一个“KeyError”：

# Defining dictionary

diction =  {

"A" :  "Z", 

"Y" :  "B",

"C" :  "X",

"W" :  "E",

"E" :  "V",

"U" :  "F",

"G" :  "T",

"S" :  "H",

"I" :  "R",

"Q" :  "J",

"K" :  "P",

"O" :  "L",

"M" :  "N",

" " :  " "

}

# Sentence in "szyfr" variable should be split into list.

szyfr = "SOME SENTENCE WHATEVER"

def split(szyfr): 

     return [char for char in szyfr]

szyfr = split(szyfr)

# Now I want to iterate through "szyfr" and replace letters as in "CAT" example:  

for i in szyfr:

        if i in diction:

                

                diction = {x:y for x,y in diction.items()}

                print(i)

                print("Variable: " + i + " is in 'key'")

                pass

        elif diction[i] in szyfr:

                diction = {y:x for x,y in diction.items()}

                print(i)

                print("Variable: " + i + " is in 'value'")

        elif i is " ":

                pass

print(szyfr)

Answer 1

您缺少一些字母，例如N。请注意，这{"M": "N"}与 不同{"N": "M"}。话虽如此，您甚至不需要字典，就好像您从 155 (65+65+26-1) 中减去大写字母的ASCII 代码（例如 A 的 65），您最终会得到 ASCII 代码对应的字母：>>> szyfr = "SOME SENTENCE WHATEVER">>> "".join(chr(155-ord(e)) if "A" <= e <= "Z" else e for e in szyfr)'HLNV HVMGVMXV DSZGVEVI'

Answer 2

根据您提供的代码，我发现以下奇怪之处：szyfr = "SOME SENTENCE WHATEVER"def split(szyfr):&nbsp;&nbsp; &nbsp; &nbsp;return [char for char in szyfr]szyfr = split(szyfr)看来您正在尝试从字符串构建一个列表，这可以简单地完成为：>>> s = "hola">>> l1 = list(s)>>> l1['h', 'o', 'l', 'a']因此，根据您的具体情况：szyfr = "SOME SENTENCE WHATEVER"szyfr = list(szyfr)不过，这并不是真正需要的，因为您可以使用for直接管理字符串，就好像它是列表一样。现在，您想要替换特定字典后面的字符。我发现你的解决方案太复杂，而你只需要：diction =&nbsp; {"A" :&nbsp; "Z",&nbsp;"Y" :&nbsp; "B","C" :&nbsp; "X","W" :&nbsp; "E","E" :&nbsp; "V","U" :&nbsp; "F","G" :&nbsp; "T","S" :&nbsp; "H","I" :&nbsp; "R","Q" :&nbsp; "J","K" :&nbsp; "P","O" :&nbsp; "L","M" :&nbsp; "N"," " :&nbsp; " "}sentence_to_code = input("Give me a sentence: ").strip().upper()toret = ""for ch in sentence_to_code:&nbsp; &nbsp; coded_ch = diction.get(ch)&nbsp; &nbsp; if not coded_ch:&nbsp; &nbsp; &nbsp; &nbsp; coded_ch = ch&nbsp; &nbsp; toret += coded_chprint(toret)如果您没有定义所有可能字符的对应项，那么使用字典的get(k)方法是明智的，该方法在未找到键k时返回None 。必须考虑到get(k)方法有默认返回值参数，以防找不到键，因此您可以使用get(k, default_return_value)，这可以让我们进一步简化代码：diction =&nbsp; {"A" :&nbsp; "Z",&nbsp;"Y" :&nbsp; "B","C" :&nbsp; "X","W" :&nbsp; "E","E" :&nbsp; "V","U" :&nbsp; "F","G" :&nbsp; "T","S" :&nbsp; "H","I" :&nbsp; "R","Q" :&nbsp; "J","K" :&nbsp; "P","O" :&nbsp; "L","M" :&nbsp; "N"," " :&nbsp; " "}sentence_to_code = input("Give me a sentence: ").strip().upper()toret = "".join([diction.get(ch, ch) for ch in sentence_to_code])print(toret)现在我们使用列表理解，因为我们不再需要条件。该调用diction.get(ch, ch)返回ch或相应的编码字符，如果在字典中找不到，则返回ch本身。通过调用str.join()，"".join(...)我们将列表转换回字符串。

Answer 3

如果您确实想使用一个字典，其中每个键“字母”都有值“相反字母”：这是一个可能的解决方案：diction = {" ": " "}all_letters = range(ord('A'), ord('Z')+1)for char, opsite_char in zip(all_letters, reversed(all_letters)):&nbsp; &nbsp; diction[chr(char)] = chr(opsite_char)print(diction)输出：{' ': ' ', 'A': 'Z', 'B': 'Y', 'C': 'X', 'D': 'W', 'E': 'V', 'F': 'U', 'G': 'T',&nbsp;'H': 'S', 'I': 'R', 'J': 'Q', 'K': 'P', 'L': 'O', 'M': 'N', 'N': 'M', 'O': 'L',&nbsp;'P': 'K', 'Q': 'J', 'R': 'I', 'S': 'H', 'T': 'G', 'U': 'F', 'V': 'E', 'W': 'D',&nbsp;'X': 'C', 'Y': 'B', 'Z': 'A'}