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How to remove a special character based on negative pattern matching using regular expression

I have a sample string of something like hello \+ \\\world \+ \\\\ this \234 \ is \Pattern\ and I want it to be something like hello + \world + this 234 is \Pattern
One way to do is to run a loop for every character in the string and if it's a \ and next character is NOT a word, then replace it with a space. Simple but inefficient code. There must be another way to do it using regular expression.
I can find all the \alphabet using r'\\\w+' and any single \ followed by space as \\\s+ but these won't take \\\ \( \+ into consideration. How can this be done?

Maybe use:
\\(?![A-Za-z])\s*
And replace with empty string as per this online demo
\\ - A backslash (escaped);
(?![A-Za-z]) - Negative lookahead to assert not being followed by alphachar;
\s* - 0+ (Greedy) whitespace-chars.

Try this regex:
\\(?=[\W\d]|$)
Substitute all the matches with an empty string
Click for Demo
Code
Explanation
\\ - matches \
(?=[\W\d]|$) - positive lookahead to make sure that the \ matched above must either be followed by a digit or a non-word or must be at the end of the string. All such matched \ are to be replaced by empty string

You can use a lookahead:
s = r"hello \+ \\\world \+ \\\\ this \234 \ is \pattern\'"
import re
s2 = re.sub(r'\\*(?![a-zA-Z])', '', s)
print(s2)
output: hello + \world + this 234 is \pattern'
How the regex works:
\\* # match any number of \
(?![a-zA-Z]) # if not followed by a letter

Python regex to replace double backslash with single backslash

I'm trying to replace all double backslashes with just a single backslash. I want to replace 'class=\\"highlight' with 'class=\"highlight'. I thought that python treats '\\' as one backslash and r'\\+' as a string with two backslashes. But when I try
In [5]: re.sub(r'\\+', '\\', string)
sre_constants.error: bogus escape (end of line)
So I tried switching the replace string with a raw string:
In [6]: re.sub(r'\\+', r'\\', string)
Out [6]: 'class=\\"highlight'
Which isn't what I need. So I tried only one backslash in the raw string:
In [7]: re.sub(r'\\+', r'\', string)
SyntaxError: EOL while scanning string literal

why not use string.replace()?
>>> s = 'some \\\\ doubles'
>>> print s
some \\ doubles
>>> print s.replace('\\\\', '\\')
some \ doubles
Or with "raw" strings:
>>> s = r'some \\ doubles'
>>> print s
some \\ doubles
>>> print s.replace('\\\\', '\\')
some \ doubles
Since the escape character is complicated, you still need to escape it so it does not escape the '

You only got one backslash in string:
>>> string = 'class=\\"highlight'
>>> print string
class=\"highlight
Now lets put another one in there
>>> string = 'class=\\\\"highlight'
>>> print string
class=\\"highlight
and then remove it again
>>> print re.sub('\\\\\\\\', r'\\', string)
class=\"highlight

Just use .replace() twice!
I had the following path: C:\\Users\\XXXXX\\Desktop\\PMI APP GIT\\pmi-app\\niton x5 test data
To convert \ to single backslashes, i just did the following:
path_to_file = path_to_file.replace('\\','*')
path_to_file = path_to_file.replace('**', '\\')
first operation creates ** for every \ and second operation escapes the first slash, replacing ** with a single \.
Result:
C:**Users**z0044wmy**Desktop**PMI APP GIT**pmi-app**GENERATED_REPORTS
C:\Users\z0044wmy\Desktop\PMI APP GIT\pmi-app\GENERATED_REPORTS

I just realized that this might be the simplest answer:
import os
os.getcwd()
The above outputs a path with \ (2 back slashes)
BUT if you wrap it with a print function, i.e.,
print(os.getcwd())
it will output the 2 slashes with 1 slash so you can then copy and paste into an address bar!

simplejson - encoding regexp \d+

I have some missunderstanding with encoding regexp:
>>> simplejson.dumps({'title':r'\d+'})
'{"title": "\\\\d+"}'
>>> simplejson.loads('{"title": "\\\\d+"}')
{u'title': u'\\d+'}
>>> print simplejson.loads('{"title": "\\\\d+"}')['title']
\d+
So, without using print I see \\, with using print I see \. So, what the value loaded dict contains - with \\ or with \?

Here is a trick: Use list to see what characters are really in the string:
In [3]: list(u'\\d+')
Out[3]: [u'\\', u'd', u'+']
list breaks up the string into individual characters. So u'\\' is one character. (The double backslash in u'\\' is an escape sequence.) It represents one backslash character. This is correct since r'\d+' also has only one backslash:
In [4]: list(r'\d+')
Out[4]: ['\\', 'd', '+']

How do I escape backslash and single quote or double quote in Python? [duplicate]

This question already has answers here:
How can I put an actual backslash in a string literal (not use it for an escape sequence)?
(4 answers)
Closed 7 months ago.
How do I escape a backslash and a single quote or double quote in Python?
For example:
Long string = '''some 'long' string \' and \" some 'escaped' strings'''
value_to_change = re.compile(A EXPRESION TO REPRESENT \' and \")
modified = re.sub(value_to_change, 'thevalue', Long_string)
## Desired Output
modified = '''some 'long' string thevalue and thevalue some 'escaped' strings'''

How you did it
If your "long string" is read from a file (as you mention in a comment) then your question is misleading. Since you obviously don't fully understand how escaping works, the question as you wrote it down probably is different from the question you really have.
If these are the contents of your file (51 bytes as shown + maybe one or two end-of-line characters):
some 'long' string \' and \" some 'escaped' strings
then this is what it will look like in python:
>>> s1 = open('data.txt', 'r').read().strip()
>>> s1
'some \'long\' string \\\' and \\" some \'escaped\' strings'
>>> print s1
some 'long' string \' and \" some 'escaped' strings
What you wrote in the question will produce:
>>> s2 = '''some 'long' string \' and \" some 'escaped' strings'''
>>> s2
'some \'long\' string \' and " some \'escaped\' strings'
>>> print s2
some 'long' string ' and " some 'escaped' strings
>>> len(s)
49
Do you see the difference?
There are no backslashes in s2 because they have special meaning when you use them to write down strings in Python. They have no special meaning when you read them from a file.
If you want to write down a string that afterwards has a backslash in it, you have to protect the backslash you enter. You have to keep Python from thinking it has special meaning. You do that by escaping it - with a backslash.
One way to do this is to use backslashes, but often the easier and less confusing way is to use raw strings:
>>> s3 = r'''some 'long' string \' and \" some 'escaped' strings'''
'some \'long\' string \\\' and \\" some \'escaped\' strings'
>>> print s3
some 'long' string \' and \" some 'escaped' strings
>>> s1 == s3
True
How you meant it
The above was only to show you that your question was confusing.
The actual answer is a bit harder - when you are working with regular expressions, the backslash takes on yet another layer of special meaning. If you want to safely get a backslash through string escaping and through regex escaping to the actual regex, you have to write down multiple backslashes accordingly.
Furthermore, rules for putting single quotes (') in single-quoted raw strings (r'') are a bit tricky as well, so I will use a raw string with triple single-quotes (r'''''').
>>> print re.sub(r'''\\['"]''', 'thevalue', s1)
some 'long' string thevalue and thevalue some 'escaped' strings
The two backslashes stay two backslashes throughout string escaping and then become only one backslash without special meaning through regex escaping. In total, the regex says:
"match one backslash followed by either a single-quote or a double-quote."
How it should be done
Now for the pièce de résistance: The previous is really a good demonstration of what jwz meant1. If you forget about regex (and know about raw strings), the solution becomes much more obvious:
>>> print s1.replace(r'\"', 'thevalue').replace(r"\'", 'thevalue')
some 'long' string thevalue and thevalue some 'escaped' strings
1 Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems.

The problem is that in your string \' and \" get converted to ' and ", so on your example as-is, you won't be able to match only \' without matching the single quotes around long.
But my understanding is that this data comes from a file so assuming you have your_file.txt containing
some 'long' string \' and \" some 'escaped' strings
you can replace \' and \" with following code:
import re
from_file = open("your_file.txt", "r").read()
print(re.sub("\\\\(\"|')", "thevalue", from_file))
Note the four slashes. Since this is a string \ gets converted to \ (as this is an escaped character). Then in the regular expression, the remaining \ gets again converted to \, as this is also regular experssion escaped character. Result will match a single slash and one of the " and ' quotes.

is this what you want?
import re
Long_string = "some long string \' and \" some escaped strings"
value_to_change = re.compile( "'|\"" )
modified = re.sub(value_to_change , 'thevalue' , Long_string )
print modified

I try this to print a single backslash (Python 3):
single_backslash_str = r'\ '[0]
print('single_backslash_str') #output: \
print('repr(single_backslash_str)') #output: '\\'
Hope this will help!

Keep in mind, all these strings are exactly the same:
Long_string = '''some long string \' and \" some escaped strings'''
Long_string = '''some long string ' and " some escaped strings'''
Long_string = """some long string ' and " some escaped strings"""
Long_string = 'some long string \' and \" some escaped strings'
Long_string = "some long string \' and \" some escaped strings"
Long_string = 'some long string \' and " some escaped strings'
Long_string = "some long string ' and \" some escaped strings"
There is no backslash character in any of them. So the regex you're looking for doesn't need to match a backslash and a quote, just a quote:
modified = re.sub("['\"]", 'thevalue', Long_string)
BTW: You also don't have to compile the regex before you use it, re.sub will accept a string regex as well as a compiled one.

Python regex - r prefix

Can anyone explain why example 1 below works, when the r prefix is not used?
I thought the r prefix must be used whenever escape sequences are used.
Example 2 and example 3 demonstrate this.
# example 1
import re
print (re.sub('\s+', ' ', 'hello there there'))
# prints 'hello there there' - not expected as r prefix is not used
# example 2
import re
print (re.sub(r'(\b\w+)(\s+\1\b)+', r'\1', 'hello there there'))
# prints 'hello there' - as expected as r prefix is used
# example 3
import re
print (re.sub('(\b\w+)(\s+\1\b)+', '\1', 'hello there there'))
# prints 'hello there there' - as expected as r prefix is not used

Because \ begin escape sequences only when they are valid escape sequences.
>>> '\n'
'\n'
>>> r'\n'
'\\n'
>>> print '\n'
>>> print r'\n'
\n
>>> '\s'
'\\s'
>>> r'\s'
'\\s'
>>> print '\s'
\s
>>> print r'\s'
\s
Unless an 'r' or 'R' prefix is present, escape sequences in strings are interpreted according to rules similar to those used by Standard C. The recognized escape sequences are:
Escape Sequence Meaning Notes
\newline Ignored
\\ Backslash (\)
\' Single quote (')
\" Double quote (")
\a ASCII Bell (BEL)
\b ASCII Backspace (BS)
\f ASCII Formfeed (FF)
\n ASCII Linefeed (LF)
\N{name} Character named name in the Unicode database (Unicode only)
\r ASCII Carriage Return (CR)
\t ASCII Horizontal Tab (TAB)
\uxxxx Character with 16-bit hex value xxxx (Unicode only)
\Uxxxxxxxx Character with 32-bit hex value xxxxxxxx (Unicode only)
\v ASCII Vertical Tab (VT)
\ooo Character with octal value ooo
\xhh Character with hex value hh
Never rely on raw strings for path literals, as raw strings have some rather peculiar inner workings, known to have bitten people in the ass:
When an "r" or "R" prefix is present, a character following a backslash is included in the string without change, and all backslashes are left in the string. For example, the string literal r"\n" consists of two characters: a backslash and a lowercase "n". String quotes can be escaped with a backslash, but the backslash remains in the string; for example, r"\"" is a valid string literal consisting of two characters: a backslash and a double quote; r"\" is not a valid string literal (even a raw string cannot end in an odd number of backslashes). Specifically, a raw string cannot end in a single backslash (since the backslash would escape the following quote character). Note also that a single backslash followed by a newline is interpreted as those two characters as part of the string, not as a line continuation.
To better illustrate this last point:
>>> r'\'
SyntaxError: EOL while scanning string literal
>>> r'\''
"\\'"
>>> '\'
SyntaxError: EOL while scanning string literal
>>> '\''
"'"
>>>
>>> r'\\'
'\\\\'
>>> '\\'
'\\'
>>> print r'\\'
\\
>>> print r'\'
SyntaxError: EOL while scanning string literal
>>> print '\\'
\

the 'r' means the the following is a "raw string", ie. backslash characters are treated literally instead of signifying special treatment of the following character.
http://docs.python.org/reference/lexical_analysis.html#literals
so '\n' is a single newline
and r'\n' is two characters - a backslash and the letter 'n'
another way to write it would be '\\n' because the first backslash escapes the second
an equivalent way of writing this
print (re.sub(r'(\b\w+)(\s+\1\b)+', r'\1', 'hello there there'))
is
print (re.sub('(\\b\\w+)(\\s+\\1\\b)+', '\\1', 'hello there there'))
Because of the way Python treats characters that are not valid escape characters, not all of those double backslashes are necessary - eg '\s'=='\\s' however the same is not true for '\b' and '\\b'. My preference is to be explicit and double all the backslashes.

Not all sequences involving backslashes are escape sequences. \t and \f are, for example, but \s is not. In a non-raw string literal, any \ that is not part of an escape sequence is seen as just another \:
>>> "\s"
'\\s'
>>> "\t"
'\t'
\b is an escape sequence, however, so example 3 fails. (And yes, some people consider this behaviour rather unfortunate.)

Try that:
a = '\''
'
a = r'\''
\'
a = "\'"
'
a = r"\'"
\'

Check below example:
print r"123\n123"
#outputs>>>
123\n123
print "123\n123"
#outputs>>>
123
123