以下是我基于Donald Miner的研究论文开发的解决此问题的方案,使用Python实现,用于应用于MAC地址的规则。
基本上,最具体的匹配来自不是任何其他匹配模式的超集的模式。针对特定问题域,您可以创建一系列测试(函数),比较两个RE并返回哪个是超集,或者它们是否正交。这使您可以构建匹配树。对于特定的输入字符串,您浏览根模式并查找任何匹配项。然后浏览其子模式。如果在任何时候正交模式匹配,则会引发错误。
设置
import re
class RegexElement:
def __init__(self, string,index):
self.string=string
self.supersets = []
self.subsets = []
self.disjoints = []
self.intersects = []
self.maybes = []
self.precompilation = {}
self.compiled = re.compile(string,re.IGNORECASE)
self.index = index
SUPERSET = object()
SUBSET = object()
INTERSECT = object()
DISJOINT = object()
EQUAL = object()
测试
每个测试都需要两个字符串(a和b),并尝试确定它们之间的关系。如果测试无法确定关系,则返回None。
SUPERSET
表示a
是b
的超集。所有与b
匹配的内容都将与a
匹配。
SUBSET
表示b
是a
的超集。
INTERSECT
表示a
的某些匹配项将与b
匹配,但某些匹配项不会,而b
的某些匹配项也不会与a
匹配。
DISJOINT
表示没有任何a
的匹配项会与b
匹配。
EQUAL
表示所有a
的匹配项都将与b
匹配,并且所有b
的匹配项都将与a
匹配。
def equal_test(a, b):
if a == b: return EQUAL
图表
class SubsetGraph(object):
def __init__(self, tests):
self.regexps = []
self.tests = tests
self._dirty = True
self._roots = None
@property
def roots(self):
if self._dirty:
r = self._roots = [i for i in self.regexps if not i.supersets]
return r
return self._roots
def add_regex(self, new_regex):
roots = self.roots
new_re = RegexElement(new_regex)
for element in roots:
self.process(new_re, element)
self.regexps.append(new_re)
def process(self, new_re, element):
relationship = self.compare(new_re, element)
if relationship:
getattr(self, 'add_' + relationship)(new_re, element)
def add_SUPERSET(self, new_re, element):
for i in element.subsets:
i.supersets.add(new_re)
new_re.subsets.add(i)
element.supersets.add(new_re)
new_re.subsets.add(element)
def add_SUBSET(self, new_re, element):
for i in element.subsets:
self.process(new_re, i)
element.subsets.add(new_re)
new_re.supersets.add(element)
def add_DISJOINT(self, new_re, element):
for i in element.subsets:
i.disjoints.add(new_re)
new_re.disjoints.add(i)
new_re.disjoints.add(element)
element.disjoints.add(new_re)
def add_INTERSECT(self, new_re, element):
for i in element.subsets:
self.process(new_re, i)
new_re.intersects.add(element)
element.intersects.add(new_re)
def add_EQUAL(self, new_re, element):
new_re.supersets = element.supersets.copy()
new_re.subsets = element.subsets.copy()
new_re.disjoints = element.disjoints.copy()
new_re.intersects = element.intersects.copy()
def compare(self, a, b):
for test in self.tests:
result = test(a.string, b.string)
if result:
return result
def match(self, text, strict=True):
matches = set()
self._match(text, self.roots, matches)
out = []
for e in matches:
for s in e.subsets:
if s in matches:
break
else:
out.append(e)
if strict and len(out) > 1:
for i in out:
print(i.string)
raise Exception("Multiple equally specific matches found for " + text)
return out
def _match(self, text, elements, matches):
new_elements = []
for element in elements:
m = element.compiled.match(text)
if m:
matches.add(element)
new_elements.extend(element.subsets)
if new_elements:
self._match(text, new_elements, matches)
用法
graph = SubsetGraph([equal_test, test_2, test_3, ...])
graph.add_regex("00:11:22:..:..:..")
graph.add_regex("..(:..){5,5}"
graph.match("00:de:ad:be:ef:00")
完整可用版本在这里。