凯撒密码，Python 3，新程序员

Question

第一次发帖到StackOverflow！

我是一个新的程序员，所以请耐心等待，因为我还没有跟上所有可能的函数，所以可能我只是目前缺少一个函数或方法的知识，可以快速解决我的问题。

感谢大家的帮助！

#string used for comparing code to
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

#strings for shifting the code
shift = "ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZ"
punctuation = ".,!;:?"

#ask user for the encrypted text
code = input("Enter text: ")

#variables storing the most common letters and the # of times they come up
common1 = 0
common2 = 0
common3 = 0
ac = 0
bc = 0
cc = 0

for i in range(0, len(alphabet)):
    # print number of times each letter comes up --- print( str(alphabet[i]) + " is present " + str(code.count(alphabet[i])) + " number of times" )

    #finding the number of times the most common letter comes up
    if code.count(alphabet[i]) > common1:
        common3 = common2
        cc = bc
        common2 = common1
        bc = ac
        common1 = code.count(alphabet[i])
        ac = alphabet[i]

    elif code.count(alphabet[i]) > common2:
        common3 = common2
        cc = bc
        common2 = code.count(alphabet[i])
        bc = alphabet[i]

    elif code.count(alphabet[i]) > common3:
        common3 = code.count(alphabet[i])
        cc = alphabet[i]


print("Most common letter, " + str(ac) + ", comes up " + str(common1) + " times")
print("Second most common letter, " + str(bc) + ", comes up " + str(common2) + " times")
print("Third most common letter, " + str(cc) + ", comes up " + str(common3) + " times")


for i in range(0, len(code)):
    a = shift.index("E") + 26 - shift.index(ac)
    print(shift[a + shift.index( code[i] ) ], end = "")

Answer 1

您应该检查每个位置的代码值是否在您的字母表中。如果不是，只需打印代码中的值，然后继续下一个循环。

for i in range(0, len(code)):
    if code[i] not in alphabet:
        print(code[i])
        continue
    a = shift.index("E") + 26 - shift.index(ac)
    print(shift[a + shift.index( code[i] ) ], end = "")