Question

我正在编写一段代码，以使用Caesar Cipher方法对邮件进行加密，如果我想对句子进行拆分（如果遇到特殊字符的话），则会弹出以下错误消息。

我尝试了其他方法，例如line.split(arg)，但是问题是它只能在我想输入数十个参数时接受一个参数，所以到目前为止，唯一的选择似乎是re.split('args [seperated by '|']', str) 。但是，

import re, textwrap
def csr():
    avalin = input('\n...\tPlease input the value of a: ')
    fcharin = input('...\tDo you want to chage foreign characters also [Y/N]? ')
    tmp = []
    strout = ''
    tmpstr = []
    if fcharin.lower() == 'n' or fcharin.lower() == 'no':
        tmpstr = (re.split('!|"|#|$|%|&|\'|(|)|*|+|,|-|.|/|:|;|<|=|>|?|@|[|\|]|^|_|`|{|||}|~|, |. |; |: |? |! | (|) | + | - | * | / | |\n|\*', strin))
    else:
        tmpstr.append(strin)
    for item in tmpstr:
        for char in item:
            tmp.append((int(avalin)+(ord(char)-97))%26)
    for item in tmp:
        strout += chr(item+97)
    wrlimit = textwrap.TextWrapper(width = 80)
    wrlist = wrlimit.wrap(text=strout)
    for line in wrlist:
        print(line)

我希望输入根据上面tmpstr中给出的字符进行拆分（应该返回一个数组），但是会弹出此错误：

  File "main.py", line 47, in <module>
    main()
  File "main.py", line 44, in main
    encrypt()
  File "C:\Users\warre\OneDrive\Desktop\School\2018-19\Semester 2\Physical Science\Science Project 2019\Science Project 2019\encrypt.py", line 55, in encrypt
    tmpstr = (re.split('!|"|#|$|%|&|\'|(|)|*|+|,|-|.|/|:|;|<|=|>|?|@|[|\|]|^|_|`|{|||}|~|, |. |; |: |? |! | (|) | + | - | * | 
/ | |\n|\*', strin))
  File "C:\Programs\Python\Python37\lib\re.py", line 213, in split
    return _compile(pattern, flags).split(string, maxsplit)
  File "C:\Programs\Python\Python37\lib\re.py", line 286, in _compile
    p = sre_compile.compile(pattern, flags)
  File "C:\Programs\Python\Python37\lib\sre_compile.py", line 764, in compile
    p = sre_parse.parse(p, flags)
  File "C:\Programs\Python\Python37\lib\sre_parse.py", line 930, in parse
    p = _parse_sub(source, pattern, flags & SRE_FLAG_VERBOSE, 0)
  File "C:\Programs\Python\Python37\lib\sre_parse.py", line 426, in _parse_sub
    not nested and not items))
  File "C:\Programs\Python\Python37\lib\sre_parse.py", line 651, in _parse
    source.tell() - here + len(this))
re.error: nothing to repeat at position 18 (line 1, column 19)

是否有更方便的拆分句子的方法，或者有使用Caesar Cipher加密消息的快捷功能吗？

Python re.error：在位置18（第1行，第19列）上没有重复的内容

0 个答案: