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Python prevent decoding HEX to ASCII while removing backslashes from my Var

I want to strip some unwanted symbols from my variable. In this case the symbols are backslashes. I am using a HEX number, and as an example I will show some short simple code down bellow. But I don't want python to convert my HEX to ASCII, how would I prevent this from happening.? I have some long shell codes for asm to work with later which are really long and removing \ by hand is a long process. I know there are different ways like using echo -e "x\x\x\x" > output etc, but my whole script will be written in python.
Thanks
>>> a = "\x31\xC0\x50\x68\x74\x76"
>>> b = a.strip("\\")
>>> print b
1�Phtv
>>> a = "\x31\x32\x33\x34\x35\x36"
>>> b = a.strip("\\")
>>> print b
123456
At the end I would like it to print my var:
>>> print b
x31x32x33x34x35x36

There are no backslashes in your variable:
>>> a = "\x31\xC0\x50\x68\x74\x76"
>>> print(a)
1ÀPhtv
Take newline for example: writing "\n" in Python will give you string with one character -- newline -- and no backslashes. See string literals docs for full syntax of these.
Now, if you really want to write string with such backslashes, you can do it with r modifier:
>>> a = r"\x31\xC0\x50\x68\x74\x76"
>>> print(a)
\x31\xC0\x50\x68\x74\x76
>>> print(a.replace('\\', ''))
x31xC0x50x68x74x76
But if you want to convert a regular string to hex-coded symbols, you can do it character by character, converting it to number ("\x31" == "1" --> 49), then to hex ("0x31"), and finally stripping the first character:
>>> a = "\x31\xC0\x50\x68\x74\x76"
>>> print(''.join([hex(ord(x))[1:] for x in a]))
'x31xc0x50x68x74x76'

There are two problems in your Code.
First the simple one:
strip() just removes one occurrence. So you should use replace("\\", ""). This will replace every backslash with "", which is the same as removing it.
The second problem is pythons behavior with backslashes:
To get your example working you need to append an 'r' in front of your string to indicate, that it is a raw string. a = r"\x31\xC0\x50\x68\x74\x76". In raw strings, a backlash doesn't escape a character but just stay a backslash.
>>> r"\x31\xC0\x50\x68\x74\x76"
'\\x31\\xC0\\x50\\x68\\x74\\x76'

In Python, how can you write the string String = "\s"? [duplicate]

This question already has answers here:
Why do backslashes appear twice?
(2 answers)
Closed 7 months ago.
Why does:
B = "The" + "\s"
and
B = "The" + r"\s"
yield:
"The\\s"
Is it possible to write the above, such that the output string is:
"The\s"
I have read similar questions on both the issue of backslashes, and their property for escaping, and the interpretation of regex characters in Python.
How to print backslash with Python?
Why can't Python's raw string literals end with a single backslash?
Does this mean there is no way to write what I want?
If it is useful, My end goal is to a write a program that adds the regex expression for space (\s) to a string where this such space:
For example, start with:
A = "The Cat and Dog"
After applying the function, this becomes:
B = "The\sCat\sand\sDog"

I believe this is related to Why does printing a tuple (list, dict, etc.) in Python double the backslashes?
The representation of the string and what it actually contains can differ.
Observe:
>>> B = "The" + "\s"
>>> B
'The\\s'
>>> print B
The\s
Furthermore
>>> A = "The Cat and Dog"
>>> B = str.replace(A, ' ', '\s')
>>> B
'The\\sCat\\sand\\sDog'
>>> print B
The\sCat\sand\sDog

From the docs:
all unrecognized escape sequences are left in the string unchanged, i.e., the backslash is left in the result
So while \s is not a proper escape sequence, Python forgives you your mistake and treats the backslash as if you had properly escaped it as \\. But when you then view the string's representation, it shows the backslash properly escaped. That said, the string only contains one backslash. It's only the representation that shows it as an escape sequence with two.

You must escape the "\"
B = "The" + "\\s"
>>> B = "The" + "\\s"
>>> print(B)
The\s
See the Escape Sequences part:
Python 3 - Lexical Analysis

Print confusion

I am new to python when i try to print "\20%" that is
>>>"\20%"
why is the shell printing '\x10%' that is, it is showing
'\x10%'
the same is happening with join also when is do
>>>l = ['test','case']
>>>"\20%".join(l)
it shows
'test\x10%case'
I am using python 2.7.3

'\20' is an octal literal, and the same as chr(2 * 8 + 0) == chr(16).
What the Python shell displays by default is not the output of print, but the representation of the given value, which is the hexadecimal '\x10'.
If you want the string \20%, you have to either escape the backaslash ('\\20%') or use a raw string literal (r'\20%'). Both will be displayed as
>>> r'\20%'
'\\20%'

\20 is an escape sequence that refers to the DLE ASCII character whose decimal value is 16 (20 in octal, 10 in hexadecimal). Such a character is printed as the \x10 hex escape by the repr function of strings.
To specify a literal \20, either double the backslash ("\\20") or use a raw string (r"\20").

Two print "\20%"
what if you print directly:
>>> print '\20%'
% # some symbol not correctly display on this page
and do using r
>>> print r'\20%'
\20%
>>> r'\20%' # what r do.
'\\20%'
>>> print '\\20%'
\20%
>>>
Some time back I had same doubt about string and I asked a question, you may find helpful

Python issue with incorrectly formated strings that contains \x

At some point our python script receives string like that:
In [1]: ab = 'asd\xeffe\ctive'
In [2]: print ab
asd�fe\ctve \ \\ \\\k\\\
Data is damaged we need escape \x to be properly interpreted as \x but \c has not special meaning in string thus must be intact.
So far the closest solution I found is do something like:
In [1]: ab = 'asd\xeffe\ctve \\ \\\\ \\\\\\k\\\\\\'
In [2]: print ab.encode('string-escape').replace('\\\\', '\\').replace("\\'", "'")
asd\xeffe\ctve \ \\ \\\k\\\
Output taken from IPython, I assumed that ab is a string not unicode string (in the later case we would have to do something like that:
def escape_string(s):
if isinstance(s, str):
s = s.encode('string-escape').replace('\\\\', '\\').replace("\\'", "'")
elif isinstance(s, unicode):
s = s.encode('unicode-escape').replace('\\\\', '\\').replace("\\'", "'")
return s

\xhh is an escape character and \x is seen as the start of this escape.

'\\' is the same as '\x5c'. It is just two different ways to write the backslash character as a Python string literal.
These literal strings: r'\c', '\\c', '\x5cc', '\x5c\x63' are identical str objects in memory.
'\xef' is a single byte (239 as an integer), but r'\xef' (same as '\\xef') is a 4-byte string: '\x5c\x78\x65\x66'.
If s[0] returns '\xef' then it is what s object actually contains. If it is wrong then fix the source of the data.
Note: string-escape also escapes \n and the like:
>>> print u'''\xef\c\\\N{SNOWMAN}"'\
... ☃\u2603\"\'\n\xa0'''.encode('unicode-escape')
\xef\\c\\\u2603"'\u2603\u2603"'\n\xa0
>>> print b'''\xef\c\\\N{SNOWMAN}"'\
... ☃\u2603\"\'\n\xa0'''.encode('string-escape')
\xef\\c\\\\N{SNOWMAN}"\'\xe2\x98\x83\\u2603"\'\n\xa0
backslashreplace is used only on characters that cause UnicodeEncodeError:
>>> print u'''\xef\c\\\N{SNOWMAN}"'\
... ☃\u2603\"\'\n\xa0'''
ï\c\☃"'☃☃"'
>>> print b'''\xef\c\\\N{SNOWMAN}"'\
... ☃\u2603\"\'\n\xa0'''
�\c\\N{SNOWMAN}"'☃\u2603"'
�
>>> print u'''\xef\c\\\N{SNOWMAN}"'\
... ☃\u2603\"\'\n\xa0'''.encode('ascii', 'backslashreplace')
\xef\c\\u2603"'\u2603\u2603"'
\xa0
>>> print b'''\xef\c\\\N{SNOWMAN}"'\
... ☃\u2603\"\'\n\xa0'''.decode('latin1').encode('ascii', 'backslashreplace')
\xef\c\\N{SNOWMAN}"'\xe2\x98\x83\u2603"'
\xa0

Backslashes introduce "escape sequences". \x specifically allows you to specify a byte, which is given as two hexadecimal digits after the x. ef are two hexadecimal digits, hence you get no error. Double the backslash to escape it, or use a raw string r"\xeffective".
Edit: While the Python console may show you '\\', this is precisely what you expect. You just say you expect something else because you confuse the string and its representation. It's a string containing a single backslash. If you were to output it with print, you'd see a single backslash.
But the string literal '\' is ill-formed (not closed because \' is an apostrophe, not a backslash and end-of-string-literal), so repr, which formats the results at the interactive shell, does not produce it. Instead it produces a string literal which you could paste into Python source code and get the same string object. For example, len('\\') == 1.

The \x escape sequence signifies a Unicode character in the string, and ef is being interpreted as the hex code. You can sanitize the string by adding an additional \, or else make it a raw string (r'\xeffective').
>>> r'\xeffective'[0]
'\\'
EDIT: You could convert an existing string using the following hack:
>>> a = '\xeffective'
>>> b = repr(a).strip("'")
>>> b
'\\xeffective'

Python Replace \\ with \ [duplicate]

This question already has answers here:
Process escape sequences in a string in Python
(8 answers)
Closed 7 months ago.
So I can't seem to figure this out... I have a string say, "a\\nb" and I want this to become "a\nb". I've tried all the following and none seem to work;
>>> a
'a\\nb'
>>> a.replace("\\","\")
File "<stdin>", line 1
a.replace("\\","\")
^
SyntaxError: EOL while scanning string literal
>>> a.replace("\\",r"\")
File "<stdin>", line 1
a.replace("\\",r"\")
^
SyntaxError: EOL while scanning string literal
>>> a.replace("\\",r"\\")
'a\\\\nb'
>>> a.replace("\\","\\")
'a\\nb'
I really don't understand why the last one works, because this works fine:
>>> a.replace("\\","%")
'a%nb'
Is there something I'm missing here?
EDIT I understand that \ is an escape character. What I'm trying to do here is turn all \\n \\t etc. into \n \t etc. and replace doesn't seem to be working the way I imagined it would.
>>> a = "a\\nb"
>>> b = "a\nb"
>>> print a
a\nb
>>> print b
a
b
>>> a.replace("\\","\\")
'a\\nb'
>>> a.replace("\\\\","\\")
'a\\nb'
I want string a to look like string b. But replace isn't replacing slashes like I thought it would.

There's no need to use replace for this.
What you have is a encoded string (using the string_escape encoding) and you want to decode it:
>>> s = r"Escaped\nNewline"
>>> print s
Escaped\nNewline
>>> s.decode('string_escape')
'Escaped\nNewline'
>>> print s.decode('string_escape')
Escaped
Newline
>>> "a\\nb".decode('string_escape')
'a\nb'
In Python 3:
>>> import codecs
>>> codecs.decode('\\n\\x21', 'unicode_escape')
'\n!'

You are missing, that \ is the escape character.
Look here: http://docs.python.org/reference/lexical_analysis.html
at 2.4.1 "Escape Sequence"
Most importantly \n is a newline character.
And \\ is an escaped escape character :D
>>> a = 'a\\\\nb'
>>> a
'a\\\\nb'
>>> print a
a\\nb
>>> a.replace('\\\\', '\\')
'a\\nb'
>>> print a.replace('\\\\', '\\')
a\nb

r'a\\nb'.replace('\\\\', '\\')
or
'a\nb'.replace('\n', '\\n')

Your original string, a = 'a\\nb' does not actually have two '\' characters, the first one is an escape for the latter. If you do, print a, you'll see that you actually have only one '\' character.
>>> a = 'a\\nb'
>>> print a
a\nb
If, however, what you mean is to interpret the '\n' as a newline character, without escaping the slash, then:
>>> b = a.replace('\\n', '\n')
>>> b
'a\nb'
>>> print b
a
b

It's because, even in "raw" strings (=strings with an r before the starting quote(s)), an unescaped escape character cannot be the last character in the string. This should work instead:
'\\ '[0]

In Python string literals, backslash is an escape character. This is also true when the interactive prompt shows you the value of a string. It will give you the literal code representation of the string. Use the print statement to see what the string actually looks like.
This example shows the difference:
>>> '\\'
'\\'
>>> print '\\'
\

In Python 3 it will be:
bytes(s, 'utf-8').decode("unicode_escape")

This works on Windows with Python 3.x:
import os
str(filepath).replace(os.path.sep, '/')
Where: os.path.sep is \ on Windows and / on Linux.
Case study
Used this to prevent errors when generating a Markdown file then rendering it to pdf.

path = "C:\\Users\\Programming\\Downloads"
# Replace \\ with a \ along with any random key multiple times
path.replace('\\', '\pppyyyttthhhooonnn')
# Now replace pppyyyttthhhooonnn with a blank string
path.replace("pppyyyttthhhooonnn", "")
print(path)
#Output...
C:\Users\Programming\Downloads