Python - 基于类型将结构过滤为结构

Question

我一直对此感到头撞了两天，尝试了各种方法，它们中的任何一个都不能以我可以使用的方式工作...

问题。

我给了一个任意的字节流。在字节中有一些语义元素。有大括号，方括号和括号。这些表明了三个不同的事情 - {}是一个不。的字节范围，例如{17}是17个字节。 []是一个字节值，例如[90:95]是字节x90，x91，x92，x93，x94，x95。 （）是字节值'OR'选项，例如（46 | 47）表示x46或x47。

还有其他的语法结构，我必须检测，“！”，“*”，“？”和“：”。

一个例子字节流：524946（46 | 58）{4} 434452367672736E

我试图对其进行过滤，所以我得到这样的：

1 string 524946 
2 token (46|58) 
3 token {4} 
4 string 434452367672736E 

一旦我把它拆了，然后我可以进一步处理它。

我来得到它的工作（其丑陋难看丑陋的代码......）最近的：http://pastebin.com/XLg2H0PW

我与一些正则表达式试过，但我能得到它不算语法单位里面的字节串作为普通的字符串元素：

range_masks_list = [(m_mask1.span()) for m_mask1 in re.finditer("\{([0-9]+|[0-9]+-[0-9]+|[0-9]+-\*)\}",sequence)] ## looks for {int}, {int-int} and {int-*} 
byte_masks_list = [(m_mask2.span()) for m_mask2 in re.finditer("\[[a-fA-F0-9]{2}:[a-fA-F0-9]{2}]",sequence)] ## looks for [a:b] where a and b are byte ranges 
options_sets_list = [(m_mask3.span()) for m_mask3 in re.finditer("\(([a-fA-F0-9]{2})+\|([a-fA-F0-9]{2})+(\|([a-fA-F0-9]{2})+)*\)",sequence)] ## looks for regex or clauses e.g. (a|b) 
string_chunk_list = [(m_mask4.span()) for m_mask4 in re.finditer("([a-fA-F0-9]{2})+",sequence)] ## looks for uninterrupted hex byte spans 

这将是这样的：

def do_fragmenter(self,sequence): 
    """ converts the grep grammer normalised string into a set of fragments and offsets for sig population""" 
    sequence = sequence.replace(" ","") 
    range_masks_list = [(m_mask1.span()) for m_mask1 in re.finditer("\{([0-9]+|[0-9]+-[0-9]+|[0-9]+-\*)\}",sequence)] ## looks for {int}, {int-int} and {int-*} 
    byte_masks_list = [(m_mask2.span()) for m_mask2 in re.finditer("\[[a-fA-F0-9]{2}:[a-fA-F0-9]{2}]",sequence)] ## looks for [a:b] where a and b are byte ranges 
    options_sets_list = [(m_mask3.span()) for m_mask3 in re.finditer("\(([a-fA-F0-9]{2})+\|([a-fA-F0-9]{2})+(\|([a-fA-F0-9]{2})+)*\)",sequence)] ## looks for regex or clauses e.g. (a|b) 
    string_chunk_list = [(m_mask4.span()) for m_mask4 in re.finditer("([a-fA-F0-9]{2})+",sequence)] ## looks for uninterupted hex byte spans 
    string_chunks = [] 
    string_chunks_len = [] 
    for pair in string_chunk_list: 
     string_chunks.append(sequence[pair[0]:pair[1]]) 
     string_chunks_len.append(len(sequence[pair[0]:pair[1]])) 
    print zip(string_chunks,string_chunks_len) 

Answer 1

只是考虑到你所定义的语法元素，你也许可以用这样的事情（用您所需要的处理打印）：

#! /usr/bin/python3.2 

import re 

a = '524946(46|58){4}434452[22:33]367672736E' 
patterns = [ ('([0-9a-fA-F]+)', 'Sequence '), 
    ('(\\([0-9a-fA-F]+\\|[0-9a-fA-F]+\\))', 'Option '), 
    ('({[0-9a-fA-F]+})', 'Curly '), 
    ('(\\[[0-9a-fA-F]+:[0-9a-fA-F]+\\])', 'Slice ') ] 

while a: 
    found = False 
    for pattern, name in patterns: 
     m = re.match (pattern, a) 
     if m: 
      m = m.groups() [0] 
      print (name + m) 
      a = a [len (m):] 
      found = True 
      break 
    if not found: raise Exception ('Unrecognized sequence') 

产量：

Sequence 524946 
Option (46|58) 
Curly {4} 
Sequence 434452 
Slice [22:33] 
Sequence 367672736E