python regex escape characters

问题

We have:

>>> str
'exit\r\ndrwxr-xr-x    2 root     root            0 Jan  1  2000 
\x1b[1;34mbin\x1b[0m\r\ndrwxr-xr-x    3 root     root           
0 Jan  1  2000 \x1b[1;34mlib\x1b[0m\r\ndrwxr-xr-x   10 root     
root            0 Jan  1  1970 \x1b[1;34mlocal\x1b[0m\r\ndrwxr-xr-x    
2 root     root            0 Jan  1  2000 \x1b[1;34msbin\x1b[0m\r\ndrwxr-xr-x    
5 root     root            0 Jan  1  2000 \x1b[1;34mshare\x1b[0m\r\n# exit\r\n'

>>> print str
exit
drwxr-xr-x    2 root     root            0 Jan  1  2000 bin
drwxr-xr-x    3 root     root            0 Jan  1  2000 lib
drwxr-xr-x   10 root     root            0 Jan  1  1970 local
drwxr-xr-x    2 root     root            0 Jan  1  2000 sbin
drwxr-xr-x    5 root     root            0 Jan  1  2000 share
# exit

I want to get rid of all the '\xblah[0m' nonsense using regexp. I've tried

re.sub(str, r'(\x.*m)', '')

But that hasn't done the trick. Any ideas?

回答1:

You have a few issues:

You're passing arguments to re.sub in the wrong order wrong. It should be:

re.sub(regexp_pattern, replacement, source_string)
The string doesn't contain "\x". That "\x1b" is the escape character, and it's a single character.
As interjay pointed out, you want ".*?" rather than ".*", because otherwise it will match everything from the first escape through the last "m".

The correct call to re.sub is:

print re.sub('\x1b.*?m', '', s)

Alternatively, you could use:

print re.sub('\x1b[^m]*m', '', s)

回答2:

You need the following changes:

Escape the backslash
Switch to non-greedy matching. Otherwise, everything between the first \x and the last m will be removed, which will be a problem when there is more than one occurrence.
The order of arguments is incorrect

Result:

re.sub(r'(\\x.*?m)', '', str)

回答3:

These are ANSI terminal codes. They're signalled by an ESC (byte 27, seen in Python as \x1B) followed by [, then some ;-separated parameters and finally a letter to specify which command it is. (m is a colour change.)

The parameters are usually numbers so for this simple case you could get rid of them with:

ansisequence= re.compile(r'\x1B\[[^A-Za-z]*[A-Za-z]')
ansisequence.sub('', string)

Technically for some (non-colour-related) control codes they could be general strings, which makes the parsing annoying. It's rare you'd meet these, but if you did I guess you'd have to use something complicated like:

\x1B\[((\d+|"[^"]*")(;(\d+|"[^"]*"))*)?[A-Za-z]

Best would be to persuade whatever's generating the string that you're not an ANSI terminal so it shouldnt include colour codes in its output.

回答4:

Try running ls --color=never -l instead, and you won't get the ANSI escape codes in the first place.

回答5:

Here is a pyparsing solution to your problem, with a general parsing expression for those pesky escape sequences. By transforming the initial string with a suppressed expression, this returns a string stripped of all matches of the expression.

s = \
'exit\r\ndrwxr-xr-x    2 root     root            0 Jan  1  2000 ' \
'\x1b[1;34mbin\x1b[0m\r\ndrwxr-xr-x    3 root     root           ' \
'0 Jan  1  2000 \x1b[1;34mlib\x1b[0m\r\ndrwxr-xr-x   10 root     ' \
'root            0 Jan  1  1970 \x1b[1;34mlocal\x1b[0m\r\ndrwxr-xr-x    ' \
'2 root     root            0 Jan  1  2000 \x1b[1;34msbin\x1b[0m\r\ndrwxr-xr-x    ' \
'5 root     root            0 Jan  1  2000 \x1b[1;34mshare\x1b[0m\r\n# exit\r\n' \

from pyparsing import (Literal, Word, nums, Combine, 
    delimitedList, oneOf, alphas, Suppress)

ESC = Literal('\x1b')
integer = Word(nums)
escapeSeq = Combine(ESC + '[' + delimitedList(integer,';') + oneOf(list(alphas)))

s_prime = Suppress(escapeSeq).transformString(s)

print s_prime

This prints your desired output, as stored in s_prime.

来源：https://stackoverflow.com/questions/1833873/python-regex-escape-characters

标签

python

regex

ansi-escape