Best way to replace \\x00 in python lists?

问题:

I have a list of values from a parsed PE file that include /x00 null bytes at the end of each section. I want to be able to remove the /x00 bytes from the string without removing all "x"s from the file. I have tried doing .replace and re.sub, but not which much success.

Using Python 2.6.6

Example.

import re

List = [['.text\x00\x00\x00'], ['.data\x00\x00\x00'], ['.rsrc\x00\x00\x00']]

while count < len(List):
    test = re.sub('\\\\x00', '', str(list[count])
    print test
    count += 1

>>>test  (removes x, but I want to keep it) #changed from tet to test
>>>data
>>>rsrc

I want to get the following output

text data rsrc

Any ideas on the best way of going about this?

回答1:

>>> L = [['.text\x00\x00\x00'], ['.data\x00\x00\x00'], ['.rsrc\x00\x00\x00']]
>>> [[x[0]] for x in L]
[['.text\x00\x00\x00'], ['.data\x00\x00\x00'], ['.rsrc\x00\x00\x00']]
>>> [[x[0].replace('\x00', '')] for x in L]
[['.text'], ['.data'], ['.rsrc']]

Or to modify the list in place instead of creating a new one:

for x in L:
    x[0] = x[0].replace('\x00', '')

回答2:

lst = (i[0].rstrip('\x00') for i in List)
for j in lst: 
   print j,

回答3:

What you're really wanting to do is replace '\x00' characters in strings in a list.

Towards that goal, people often overlook the fact that in Python 2 the non-Unicode string translate() method will also optionally (or only) delete 8-bit characters as illustrated below. (It doesn't accept this argument in Python 3 because strings are Unicode objects by default.)

Your List data structure seems a little odd, since it's a list of one-element lists consisting of just single strings. Regardless, in the code below I've renamed it sections since Capitalized words should only be used for the names of classes according to PEP 8 -- Style Guide for Python Code.

sections = [['.text\x00\x00\x00'], ['.data\x00\x00\x00'], ['.rsrc\x00\x00\x00']]

for section in sections:
    test = section[0].translate(None, '\x00')
    print test

Output:

.text
.data
.rsrc

回答4:

Try a unicode pattern, like this:

re.sub(u'\x00', '', s)

It should give the following results:

l = [['.text\x00\x00\x00'], ['.data\x00\x00\x00'], ['.rsrc\x00\x00\x00']]
for x in l:
    for s in l:
        print re.sub(u'\x00', '', s)
        count += 1

.text
.data
.rsrc

Or, using list comprehensions:

[[re.sub(u'\x00', '', s) for s in x] for x in l]

Actually, should work without the 'u' in front of the string. Just remove the first 3 slashes, and use this as your regex pattern:

'\x00'

回答5:

I think a better way to take care of this particular problem is to use the following function:

import string

for item  in List:
  filter(lambda x: x in string.printable, str(item))

This will get rid of not just \x00 but any other such hex values that are appended to your string.

回答6:

from itertools import chain

List = [['.text\x00\x00\x00'], ['.data\x00\x00\x00'], ['.rsrc\x00\x00\x00']]    
new_list = [x.replace("\x00", "") for x in chain(*List)]
#['.text', '.data', '.rsrc']