您可以在这里使用^{}。您仍然需要扫描字符串，但正则表达式也需要：from itertools import groupby
three_or_more = (char for char, group in groupby(input_string)
if sum(1 for _ in group) >= 3)
这将生成一个生成器；对其进行迭代以列出3次或更多次找到的所有字符，或者使用any()查看是否至少有一个这样的组：
^{pr2}$
不幸的是，re解决方案在这里更有效：>>> from timeit import timeit
>>> import random
>>> from itertools import groupby
>>> import re
>>> import string
>>> def consecutive_groupby(string):
... three_or_more = (char for char, group in groupby(string)
... if sum(1 for _ in group) >= 3)
... return any(three_or_more)
...
>>> def consecutive_re(string):
... return re.search(r'(.)\1\1', string) is not None
...
>>> # worst-case: random data with no consecutive strings
...
>>> test_string = ''.join([random.choice(string.ascii_letters) for _ in range(1000)])
>>> consecutive_re(test_string), consecutive_groupby(test_string)
(False, False)
>>> timeit('consecutive(s)', 'from __main__ import test_string as s, consecutive_re as consecutive', number=10000)
0.19730806350708008
>>> timeit('consecutive(s)', 'from __main__ import test_string as s, consecutive_groupby as consecutive', number=10000)
4.633949041366577
>>> # insert repeated characters
...
>>> test_string_with_repeat = test_string[:100] + 'aaa' + test_string[100:]
>>> consecutive_re(test_string_with_repeat), consecutive_groupby(test_string_with_repeat)
(True, True)
>>> timeit('consecutive(s)', 'from __main__ import test_string_with_repeat as s, consecutive_re as consecutive', number=10000)
0.03344106674194336
>>> timeit('consecutive(s)', 'from __main__ import test_string_with_repeat as s, consecutive_groupby as consecutive', number=10000)
0.4827418327331543
Avinash给出的正则表达式方法是这里明显的赢家，这表明您应该始终度量替代方案。在