[假设您需要这个，因为它是一些需要regexp的奇怪的第三方系统]
我对弗雷德里克的评论的思考越多，我就越同意。即使输入字符串很长，regexp引擎也应该能够将其编译为紧凑的DFA。在许多情况下，以下是一个明智的解决方案：
import re
def regexp(lo, hi):
fmt = '%%0%dd' % len(str(hi))
return re.compile('(%s)' % '|'.join(fmt % i for i in range(lo, hi+1)))
（它适用于以下测试中的所有数值范围，包括99519000-99519099。粗略的信封计算表明9位数字约为1GB内存的限制。匹配；如果只有几个匹配，则可以更大。）
[再次更新以提供更短的结果-除了合并偶尔出现\d\d的效果与手工生成的效果差不多之外]
假设所有数字的长度相同（即，如有必要，则在左侧使用零填充），这可以实现：
import re
def alt(*args):
'''format regexp alternatives'''
if len(args) == 1: return args[0]
else: return '(%s)' % '|'.join(args)
def replace(s, c):
'''replace all characters in a string with a different character'''
return ''.join(map(lambda x: c, s))
def repeat(s, n):
'''format a regexp repeat'''
if n == 0: return ''
elif n == 1: return s
else: return '%s{%d}' % (s, n)
def digits(lo, hi):
'''format a regexp digit range'''
if lo == 0 and hi == 9: return r'\d'
elif lo == hi: return str(lo)
else: return '[%d-%d]' % (lo, hi)
def trace(f):
'''for debugging'''
def wrapped(lo, hi):
result = f(lo, hi)
print(lo, hi, result)
return result
return wrapped
#@trace # uncomment to get calls traced to stdout (explains recursion when bug hunting)
def regexp(lo, hi):
'''generate a regexp that matches integers from lo to hi only.
assumes that inputs are zero-padded to the length of hi (like phone numbers).
you probably want to surround with ^ and $ before using.'''
assert lo <= hi
assert lo >= 0
slo, shi = str(lo), str(hi)
# zero-pad to same length
while len(slo) < len(shi): slo = '0' + slo
# first digits and length
l, h, n = int(slo[0]), int(shi[0]), len(slo)
if l == h:
# extract common prefix
common = ''
while slo and slo[0] == shi[0]:
common += slo[0]
slo, shi = slo[1:], shi[1:]
if slo: return common + regexp(int(slo), int(shi))
else: return common
else:
# the core of the routine.
# split into 'complete blocks' like 200-599 and 'edge cases' like 123-199
# and handle each separately.
# are these complete blocks?
xlo = slo[1:] == replace(slo[1:], '0')
xhi = shi[1:] == replace(shi[1:], '9')
# edges of possible complete blocks
mlo = int(slo[0] + replace(slo[1:], '9'))
mhi = int(shi[0] + replace(shi[1:], '0'))
if xlo:
if xhi:
# complete block on both sides
# this is where single digits are finally handled, too.
return digits(l, h) + repeat('\d', n-1)
else:
# complete block to mhi, plus extra on hi side
prefix = '' if l or h-1 else '0'
return alt(prefix + regexp(lo, mhi-1), regexp(mhi, hi))
else:
prefix = '' if l else '0'
if xhi:
# complete block on hi side plus extra on lo
return alt(prefix + regexp(lo, mlo), regexp(mlo+1, hi))
else:
# neither side complete, so add extra on both sides
# (and maybe a complete block in the middle, if room)
if mlo + 1 == mhi:
return alt(prefix + regexp(lo, mlo), regexp(mhi, hi))
else:
return alt(prefix + regexp(lo, mlo), regexp(mlo+1, mhi-1), regexp(mhi, hi))
# test a bunch of different ranges
for (lo, hi) in [(0, 0), (0, 1), (0, 2), (0, 9), (0, 10), (0, 11), (0, 101),
(1, 1), (1, 2), (1, 9), (1, 10), (1, 11), (1, 101),
(0, 123), (111, 123), (123, 222), (123, 333), (123, 444),
(0, 321), (111, 321), (222, 321), (321, 333), (321, 444),
(123, 321), (111, 121), (121, 222), (1234, 4321), (0, 999),
(99519000, 99519099)]:
fmt = '%%0%dd' % len(str(hi))
rx = regexp(lo, hi)
print('%4s - %-4s %s' % (fmt % lo, fmt % hi, rx))
m = re.compile('^%s$' % rx)
for i in range(0, 1+int(replace(str(hi), '9'))):
if m.match(fmt % i):
assert lo <= i <= hi, i
else:
assert i < lo or i > hi, i
该函数regexp(lo, hi)会建立一个正则表达式，以匹配介于lo和之间的值hi（零填充到最大长度）。您可能需要在^前后放置一个$字符串（如测试代码中所示），以将匹配项强制为整个字符串。
该算法实际上非常简单- 它以递归方式将事物划分为通用前缀和“完整块”。完整的区块类似于200-599，并且可以可靠地匹配（在这种情况下为[2-5]\d{2}）。
因此123-599被分为123-199和200-599。后半部分是一个完整的块，上半部分有一个公共前缀1和23-99，将其递归处理为23-29（公共前缀）和30-99（完整块）（我们最终终止，因为参数每个通话的时间都比初始输入短）。
唯一令人讨厌的细节是prefix，这是必需的，因为to的参数regexp()是整数，因此在调用生成00-09的regexp时，实际上会生成0-9的regexp，而前导0除外。
0 - 0 0
0 - 1 [0-1]
0 - 2 [0-2]
0 - 9 \d
00 - 10 (0\d|10)
00 - 11 (0\d|1[0-1])
000 - 101 (0\d\d|10[0-1])
1 - 1 1
1 - 2 [1-2]
1 - 9 [1-9]
01 - 10 (0[1-9]|10)
01 - 11 (0[1-9]|1[0-1])
001 - 101 (0(0[1-9]|[1-9]\d)|10[0-1])
000 - 123 (0\d\d|1([0-1]\d|2[0-3]))
111 - 123 1(1[1-9]|2[0-3])
123 - 222 (1(2[3-9]|[3-9]\d)|2([0-1]\d|2[0-2]))
123 - 333 (1(2[3-9]|[3-9]\d)|2\d\d|3([0-2]\d|3[0-3]))
123 - 444 (1(2[3-9]|[3-9]\d)|[2-3]\d{2}|4([0-3]\d|4[0-4]))
000 - 321 ([0-2]\d{2}|3([0-1]\d|2[0-1]))
111 - 321 (1(1[1-9]|[2-9]\d)|2\d\d|3([0-1]\d|2[0-1]))
222 - 321 (2(2[2-9]|[3-9]\d)|3([0-1]\d|2[0-1]))
321 - 333 3(2[1-9]|3[0-3])
321 - 444 (3(2[1-9]|[3-9]\d)|4([0-3]\d|4[0-4]))
123 - 321 (1(2[3-9]|[3-9]\d)|2\d\d|3([0-1]\d|2[0-1]))
111 - 121 1(1[1-9]|2[0-1])
121 - 222 (1(2[1-9]|[3-9]\d)|2([0-1]\d|2[0-2]))
1234 - 4321 (1(2(3[4-9]|[4-9]\d)|[3-9]\d{2})|[2-3]\d{3}|4([0-2]\d{2}|3([0-1]\d|2[0-1])))
000 - 999 \d\d{2}
99519000 - 99519099 995190\d\d
最后的测试循环超过99999999个数字需要一段时间。
表达式应该足够紧凑以避免任何缓冲区限制（我想最坏情况下的内存大小与最大数位的平方成正比）。
PS：我使用的是python 3，但我认为在这里没有太大的区别。
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