在Python中有效匹配多个正则表达式

Question

当你有正则表达式时，词法分析器很容易编写。今天我想用Python编写一个简单的通用分析器，并提出：

import re
import sys

class Token(object):
    """ A simple Token structure.
        Contains the token type, value and position. 
    """
    def __init__(self, type, val, pos):
        self.type = type
        self.val = val
        self.pos = pos

    def __str__(self):
        return '%s(%s) at %s' % (self.type, self.val, self.pos)


class LexerError(Exception):
    """ Lexer error exception.

        pos:
            Position in the input line where the error occurred.
    """
    def __init__(self, pos):
        self.pos = pos


class Lexer(object):
    """ A simple regex-based lexer/tokenizer.

        See below for an example of usage.
    """
    def __init__(self, rules, skip_whitespace=True):
        """ Create a lexer.

            rules:
                A list of rules. Each rule is a `regex, type`
                pair, where `regex` is the regular expression used
                to recognize the token and `type` is the type
                of the token to return when it's recognized.

            skip_whitespace:
                If True, whitespace (\s+) will be skipped and not
                reported by the lexer. Otherwise, you have to 
                specify your rules for whitespace, or it will be
                flagged as an error.
        """
        self.rules = []

        for regex, type in rules:
            self.rules.append((re.compile(regex), type))

        self.skip_whitespace = skip_whitespace
        self.re_ws_skip = re.compile('\S')

    def input(self, buf):
        """ Initialize the lexer with a buffer as input.
        """
        self.buf = buf
        self.pos = 0

    def token(self):
        """ Return the next token (a Token object) found in the 
            input buffer. None is returned if the end of the 
            buffer was reached. 
            In case of a lexing error (the current chunk of the
            buffer matches no rule), a LexerError is raised with
            the position of the error.
        """
        if self.pos >= len(self.buf):
            return None
        else:
            if self.skip_whitespace:
                m = self.re_ws_skip.search(self.buf[self.pos:])

                if m:
                    self.pos += m.start()
                else:
                    return None

            for token_regex, token_type in self.rules:
                m = token_regex.match(self.buf[self.pos:])

                if m:
                    value = self.buf[self.pos + m.start():self.pos + m.end()]
                    tok = Token(token_type, value, self.pos)
                    self.pos += m.end()
                    return tok

            # if we're here, no rule matched
            raise LexerError(self.pos)

    def tokens(self):
        """ Returns an iterator to the tokens found in the buffer.
        """
        while 1:
            tok = self.token()
            if tok is None: break
            yield tok


if __name__ == '__main__':
    rules = [
        ('\d+',             'NUMBER'),
        ('[a-zA-Z_]\w+',    'IDENTIFIER'),
        ('\+',              'PLUS'),
        ('\-',              'MINUS'),
        ('\*',              'MULTIPLY'),
        ('\/',              'DIVIDE'),
        ('\(',              'LP'),
        ('\)',              'RP'),
        ('=',               'EQUALS'),
    ]

    lx = Lexer(rules, skip_whitespace=True)
    lx.input('erw = _abc + 12*(R4-623902)  ')

    try:
        for tok in lx.tokens():
            print tok
    except LexerError, err:
        print 'LexerError at position', err.pos

它运作得很好，但我有点担心它太低效了。是否有任何正则表达式技巧可以让我以更有效/更优雅的方式编写它？

具体来说，有没有办法避免线性地遍历所有正则表达式规则以找到适合的规则？

Answer 1

您可以使用<！>“; <！>”将所有正则表达式合并为一个运算符，让正则表达式库执行标记之间的辨别工作。应该注意确保令牌的首选项（例如，避免将关键字作为标识符进行匹配）。

Answer 2

我建议使用re.Scanner类，它没有在标准库中记录，但它非常值得使用。这是一个例子：

import re

scanner = re.Scanner([
    (r"-?[0-9]+\.[0-9]+([eE]-?[0-9]+)?", lambda scanner, token: float(token)),
    (r"-?[0-9]+", lambda scanner, token: int(token)),
    (r" +", lambda scanner, token: None),
])

>>> scanner.scan("0 -1 4.5 7.8e3")[0]
[0, -1, 4.5, 7800.0]

Answer 3

我在python文档中找到了这个。它简单而优雅。

import collections
import re

Token = collections.namedtuple('Token', ['typ', 'value', 'line', 'column'])

def tokenize(s):
    keywords = {'IF', 'THEN', 'ENDIF', 'FOR', 'NEXT', 'GOSUB', 'RETURN'}
    token_specification = [
        ('NUMBER',  r'\d+(\.\d*)?'), # Integer or decimal number
        ('ASSIGN',  r':='),          # Assignment operator
        ('END',     r';'),           # Statement terminator
        ('ID',      r'[A-Za-z]+'),   # Identifiers
        ('OP',      r'[+*\/\-]'),    # Arithmetic operators
        ('NEWLINE', r'\n'),          # Line endings
        ('SKIP',    r'[ \t]'),       # Skip over spaces and tabs
    ]
    tok_regex = '|'.join('(?P<%s>%s)' % pair for pair in token_specification)
    get_token = re.compile(tok_regex).match
    line = 1
    pos = line_start = 0
    mo = get_token(s)
    while mo is not None:
        typ = mo.lastgroup
        if typ == 'NEWLINE':
            line_start = pos
            line += 1
        elif typ != 'SKIP':
            val = mo.group(typ)
            if typ == 'ID' and val in keywords:
                typ = val
            yield Token(typ, val, line, mo.start()-line_start)
        pos = mo.end()
        mo = get_token(s, pos)
    if pos != len(s):
        raise RuntimeError('Unexpected character %r on line %d' %(s[pos], line))

statements = '''
    IF quantity THEN
        total := total + price * quantity;
        tax := price * 0.05;
    ENDIF;
'''

for token in tokenize(statements):
    print(token)

这里的诀窍是：

tok_regex = '|'.join('(?P<%s>%s)' % pair for pair in token_specification)

此处(?P<ID>PATTERN)将使用ID指定的名称标记匹配的结果。

Answer 4

re.match被锚定。你可以给它一个位置参数：

pos = 0
end = len(text)
while pos < end:
    match = regexp.match(text, pos)
    # do something with your match
    pos = match.end()

查看带有大量词法分析器的pygments，用于不同实现的语法高亮显示，大多数基于正则表达式。

Answer 5

组合令牌正则表达式可能会有效，但您必须对其进行基准测试。类似的东西：

x = re.compile('(?P<NUMBER>[0-9]+)|(?P<VAR>[a-z]+)')
a = x.match('9999').groupdict() # => {'VAR': None, 'NUMBER': '9999'}
if a:
    token = [a for a in a.items() if a[1] != None][0]

过滤器是您必须进行基准测试的地方......

更新：我对此进行了测试，看起来好像是如上所述组合所有令牌并编写如下函数：

def find_token(lst):
    for tok in lst:
        if tok[1] != None: return tok
    raise Exception

你可以获得大致相同的速度（可能快一点）。我认为加速必须是匹配的调用次数，但令牌识别的循环仍然存在，这当然会导致它。

Answer 6

这不是您问题的直接答案，但您可能需要查看 ANTLR 。根据这个文档，python代码生成目标应该是up到目前为止。

至于你的正则表达式，如果你坚持使用正则表达式，有两种方法可以加快速度。第一种方法是按照在默认文本中找到它们的概率顺序来排序正则表达式。您可以想象添加一个简单的探查器到代码，该代码收集每个令牌类型的令牌计数并在工作体上运行词法分析器。另一个解决方案是对正则表达式进行分类（因为您的密钥空间是一个字符，相对较小），然后在对第一个字符执行单一区分后使用数组或字典来执行所需的正则表达式。

但是，我认为如果你要走这条路，你应该尝试像 ANTLR 这将更容易维护，更快，并且不太可能有bug。