在Python中有效地匹配多个regex

Question

当有正则表达式时，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，并得出：

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

它工作得很好，但我有点担心它效率太低。有没有什么regex技巧可以让我以更高效/优雅的方式编写它？

具体来说，有没有一种方法可以避免在所有的regex规则上线性循环以找到一个合适的规则？

Answer 1

我建议使用re.Scanner类，它没有在标准库中记录，但很值得使用。下面是一个例子：

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 2

您可以使用“|”运算符将所有regex合并为一个regex，并让regex库完成区分标记的工作。应注意确保标记的首选项（例如，避免将关键字匹配为标识符）。

Answer 3

我在python文档中找到了this。只是简单而优雅。

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)

这里(?P<ID>PATTERN)将用ID指定的名称标记匹配的结果。