>>> info = lambda data: len(data)>10 and data[:10]+'...' or data
       >>> info('sdfsdfsdfsdfsdfsdfsdfsdfsdfsdfsdf')
           'sdfsdfsdfs...'
       >>> info('sdfsdf')
           'sdfsdf'
       >>> 

如果你想做一些更复杂的字符串截断，你可以采用sklearn方法作为实现:

sklearn.base.BaseEstimator.__repr__ (参见原始完整代码:https://github.com/scikit-learn/scikit-learn/blob/f3f51f9b6/sklearn/base.py#L262)

它增加了一些好处，比如避免在单词中间截断。

def truncate_string(data, N_CHAR_MAX=70):
    # N_CHAR_MAX is the (approximate) maximum number of non-blank
    # characters to render. We pass it as an optional parameter to ease
    # the tests.

    lim = N_CHAR_MAX // 2  # apprx number of chars to keep on both ends
    regex = r"^(\s*\S){%d}" % lim
    # The regex '^(\s*\S){%d}' % n
    # matches from the start of the string until the nth non-blank
    # character:
    # - ^ matches the start of string
    # - (pattern){n} matches n repetitions of pattern
    # - \s*\S matches a non-blank char following zero or more blanks
    left_lim = re.match(regex, data).end()
    right_lim = re.match(regex, data[::-1]).end()
    if "\n" in data[left_lim:-right_lim]:
        # The left side and right side aren't on the same line.
        # To avoid weird cuts, e.g.:
        # categoric...ore',
        # we need to start the right side with an appropriate newline
        # character so that it renders properly as:
        # categoric...
        # handle_unknown='ignore',
        # so we add [^\n]*\n which matches until the next \n
        regex += r"[^\n]*\n"
        right_lim = re.match(regex, data[::-1]).end()
    ellipsis = "..."
    if left_lim + len(ellipsis) < len(data) - right_lim:
        # Only add ellipsis if it results in a shorter repr
        data = data[:left_lim] + "..." + data[-right_lim:]
    return data

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

刚刚收到的消息:

n = 8
s = '123'
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]
s = '12345678'
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]
s = '123456789'     
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]
s = '123456789012345'
print  s[:n-3] + (s[n-3:], '...')[len(s) > n]

123
12345678
12345...
12345...

limit = 75
info = data[:limit] + '..' * (len(data) > limit)

这是另一种解决方案。在测试结束时，你会得到一些关于测试的反馈。

data = {True: data[:75] + '..', False: data}[len(data) > 75]