这个方法不使用任何if:

数据[:75]+布尔(数据[75:])*'.'

info = data[:75] + ('..' if len(data) > 75 else '')

info = data[:min(len(data), 75)

假设string是我们希望截断的字符串，而nchars是输出字符串中所需的字符数。

stryng = "sadddddddddddddddddddddddddddddddddddddddddddddddddd"
nchars = 10

我们可以像下面这样截断字符串:

def truncate(stryng:str, nchars:int):
    return (stryng[:nchars - 6] + " [...]")[:min(len(stryng), nchars)]

某些测试用例的结果如下所示:

s = "sadddddddddddddddddddddddddddddd!"
s = "sa" + 30*"d" + "!"

truncate(s, 2)                ==  sa
truncate(s, 4)                ==  sadd
truncate(s, 10)               ==  sadd [...]
truncate(s, len(s)//2)        ==  sadddddddd [...]

我的解决方案为上面的测试用例产生了合理的结果。

但一些病理病例如下:

一些病理病例!

truncate(s, len(s) - 3)()       ==  sadddddddddddddddddddddd [...]
truncate(s, len(s) - 2)()       ==  saddddddddddddddddddddddd [...]
truncate(s, len(s) - 1)()       ==  sadddddddddddddddddddddddd [...]
truncate(s, len(s) + 0)()       ==  saddddddddddddddddddddddddd [...]
truncate(s, len(s) + 1)()       ==  sadddddddddddddddddddddddddd [...
truncate(s, len(s) + 2)()       ==  saddddddddddddddddddddddddddd [..
truncate(s, len(s) + 3)()       ==  sadddddddddddddddddddddddddddd [.
truncate(s, len(s) + 4)()       ==  saddddddddddddddddddddddddddddd [
truncate(s, len(s) + 5)()       ==  sadddddddddddddddddddddddddddddd 
truncate(s, len(s) + 6)()       ==  sadddddddddddddddddddddddddddddd!
truncate(s, len(s) + 7)()       ==  sadddddddddddddddddddddddddddddd!
truncate(s, 9999)()             ==  sadddddddddddddddddddddddddddddd!

值得注意的是,

当字符串包含换行字符(\n)时，可能会出现问题。 当nchars > len(s)时，我们应该打印字符串s，而不是试图打印“[…]”

下面是更多的代码:

import io

class truncate:
    """
        Example of Code Which Uses truncate:
        ```
            s = "\r<class\n 'builtin_function_or_method'>"
            s = truncate(s, 10)()
            print(s)
                    ```
                Examples of Inputs and Outputs:
                        truncate(s, 2)()   ==  \r
                        truncate(s, 4)()   ==  \r<c
                        truncate(s, 10)()  ==  \r<c [...]
                        truncate(s, 20)()  ==  \r<class\n 'bu [...]
                        truncate(s, 999)() ==  \r<class\n 'builtin_function_or_method'>
                    ```
                Other Notes:
                    Returns a modified copy of string input
                    Does not modify the original string
            """
    def __init__(self, x_stryng: str, x_nchars: int) -> str:
        """
        This initializer mostly exists to sanitize function inputs
        """
        try:
            stryng = repr("".join(str(ch) for ch in x_stryng))[1:-1]
            nchars = int(str(x_nchars))
        except BaseException as exc:
            invalid_stryng =  str(x_stryng)
            invalid_stryng_truncated = repr(type(self)(invalid_stryng, 20)())

            invalid_x_nchars = str(x_nchars)
            invalid_x_nchars_truncated = repr(type(self)(invalid_x_nchars, 20)())

            strm = io.StringIO()
            print("Invalid Function Inputs", file=strm)
            print(type(self).__name__, "(",
                  invalid_stryng_truncated,
                  ", ",
                  invalid_x_nchars_truncated, ")", sep="", file=strm)
            msg = strm.getvalue()

            raise ValueError(msg) from None

        self._stryng = stryng
        self._nchars = nchars

    def __call__(self) -> str:
        stryng = self._stryng
        nchars = self._nchars
        return (stryng[:nchars - 6] + " [...]")[:min(len(stryng), nchars)]

来的很晚，我想添加我的解决方案，以修剪文本在字符级别，也处理空白适当。

def trim_string(s: str, limit: int, ellipsis='…') -> str:
    s = s.strip()
    if len(s) > limit:
        return s[:limit-1].strip() + ellipsis
    return s

简单，但它将确保limit=6的hello world不会导致一个丑陋的hello…，而是hello…。

它还删除开头和结尾的空格，但不删除里面的空格。如果你也想删除里面的空格，签出这篇stackoverflow文章

更简明的说:

data = data[:75]

如果小于75个字符，则不会有任何更改。