python3 replacing double backslash with single backslash [duplicate]

Answer 1

If you do

s  = '\\xa5\\xc0\\xe6aK\\xf9\\x80\\xb1\\xc8*\x01\x12$\\xfbp\x1e(4\\xd6{;Z\\x'

s = s.replace('\\','\')

print(s)

you get

 File "main.py", line 3
    s = s.replace('\\','\')
                         ^
SyntaxError: EOL while scanning string literal

because in '\' the \ escapes the ' . Your string is left open.

You do not have any double \ in s - its just displaying it as such, do distinguish it from \ used to escape stuff if you inspect it.

If you print(s) you get \xa5\xc0\xe6aK\xf9\x80\xb1\xc8*$\xfbp(4\xd6{;Z\x

Answer 2

Take a closer look at the string, they are all single slash.

In [26]: my_str[0]
Out[26]: '\\'

In [27]: my_str[1]
Out[27]: 'x'

In [28]: len(my_str[0])
Out[28]: 1

And my_str.replace('\\','\') won't work because the token here is \', which escapes ' and waits for the another closing '.
Use my_str.replace('\\', '') instead

---

Update: after few more days, I realize the following discussion may also be helpful. If the intension of a string with escape ('\\x' or '\\u') are eventually hex/unicode literals, they can be decoded by escape_decode.

import codecs
print(len(b'\x32'), b'\x32')                # 1 hex literal, '\x32' == '2'
print(len(b'\\x32'), b'\\x32')              # 4 chars including escapes
print(codecs.escape_decode('\\x32', 'hex')) # chars->literal, 4->1

# 1 b'2'
# 4 b'\\x32'
# (b'2', 4)

s = '\\xa5\\xc0\\xe6aK\\xf9\\x80\\xb1\\xc8*\x01\x12$\\xfbp\x1e(4\\xd6{;Z'
ed, _ = codecs.escape_decode(s, 'hex')
print(len(s), s)
print(len(ed), ed)

# 49 \xa5\xc0\xe6aK\xf9\x80\xb1\xc8*$\xfbp(4\xd6{;Z
# 22 b'\xa5\xc0\xe6aK\xf9\x80\xb1\xc8*\x01\x12$\xfbp\x1e(4\xd6{;Z'