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Delete specific duplicated punctuation from string

I have this string s = "(0|\\+33)[1-9]( *[0-9]{2}){4}". And I want to delete just the duplicated just one ' \ ', like I want the rsult to look like (0|\+33)[1-9]( *[0-9]{2}){4}.
When I used this code, all the duplicated characters are removed:
result = "".join(dict.fromkeys(s)).
But in my case I want just to remove the duplicated ' \ '. Any help is highly appreciated

A solution using the re module:
import re
s = r"(0|\\+33)[1-9]( *[0-9]{2}){4}"
s = re.sub(r"\\(?=\\)", "", s)
print(s)
I look for all backslashes, that are followed by another backslash and replace it with an empty sign.
Output: (0|\+33)[1-9]( *[0-9]{2}){4}​

The function you need is replace
s = "(0|\\+33)[1-9]( *[0-9]{2}){4}"
result = s.replace("\\","")
EDIT
I see now that you want to remove just one \ and not both.
In order to do this you have to modify the call to replace this way
result = s.replace("\","",1) # last argument is the number of occurrances to replace
or
result = s.replace("\\","\")
EDIT of the EDIT
Backslashes are special in Python.
I'm using Python 3.10.5. If I do
x = "ab\c"
y = "ab\\c"
print(len(x)==len(y))
I get a True.
That's because backslashes are used to escape special characters, and that makes the backslash a special character :)
I suggest you to try a little bit with replace until you get what you need.

regex to extract data between quotes

As title says string is '="24digit number"' and I want to extract number between "" (example: ="000021484123647598423458" should get me '000021484123647598423458').
There are answers that answer how to get data between " but in my case I also need to confirm that =" exist without capturing (there are also other "\d{24}" strings, but they are for other stuff) it.
I couldn't modify these answers to get what I need.
My latest regex was ((?<=\")\d{24}(?=\")) and string is ="000021484123647598423458".
UPDATE: I think I will settle with pattern r'^(?:\=\")(\d{24})(?:\")' because I just want to capture digit characters.
word = '="000021484123647598423458"'
pattern = r'^(?:\=\")(\d{24})(?:\")'
match = re.findall(pattern, word)[0]
Thank you all for suggestions.

You could have it like:
=(['"])(\d{24})\1
See a demo on regex101.com.
In Python:
import re
string = '="000021484123647598423458"'
rx = re.compile(r'''=(['"])(\d{24})\1''')
print(rx.search(string).group(2))
# 000021484123647598423458

Any one of the following works:
>>> st = '="000021484123647598423458"'
>>> import re
>>> re.findall(r'".*\d+.*"',st)
['"000021484123647598423458"']
or
>>> re.findall(r'".*\d{24}.*"',st)
['"000021484123647598423458"']
or
>>> re.findall(r'"\d{24}"',st)
['"000021484123647598423458"']

How to split a string and keeping the pattern

This is how the string splitting works for me right now:
output = string.encode('UTF8').split('}/n}')[0]
output += '}\n}'
But I am wondering if there is a more pythonic way to do it.
The goal is to get everything before this '}/n}' including '}/n}'.

This might be a good use of str.partition.
string = '012za}/n}ddfsdfk'
parts = string.partition('}/n}')
# ('012za', '}/n}', 'ddfsdfk')
''.join(parts[:-1])
# 012za}/n}
Or, you can find it explicitly with str.index.
repl = '}/n}'
string[:string.index(repl) + len(repl)]
# 012za}/n}
This is probably better than using str.find since an exception will be raised if the substring isn't found, rather than producing nonsensical results.
It seems like anything "more elegant" would require regular expressions.
import re
re.search('(.*?}/n})', string).group(0)
# 012za}/n}

It can be done with with re.split() -- the key is putting parens around the split pattern to preserve what you split on:
import re
output = "".join(re.split(r'(}/n})', string.encode('UTF8'))[:2])
However, I doubt that this is either the most efficient nor most Pythonic way to achieve what you want. I.e. I don't think this is naturally a split sort of problem. For example:
tag = '}/n}'
encoded = string.encode('UTF8')
output = encoded[:encoded.index(tag)] + tag
or if you insist on a one-liner:
output = (lambda string, tag: string[:string.index(tag)] + tag)(string.encode('UTF8'), '}/n}')
or returning to regex:
output = re.match(r".*}/n}", string.encode('UTF8')).group(0)

>>> string_to_split = 'first item{\n{second item'
>>> sep = '{\n{'
>>> output = [item + sep for item in string_to_split.split(sep)]
NOTE: output = ['first item{\n{', 'second item{\n{']
then you can use the result:
for item_with_delimiter in output:
...
It might be useful to look up os.linesep if you're not sure what the line ending will be. os.linesep is whatever the line ending is under your current OS, so '\r\n' under Windows or '\n' under Linux or Mac. It depends where input data is from, and how flexible your code needs to be across environments.

Adapted from Slice a string after a certain phrase?, you can combine find and slice to get the first part of the string and retain }/n}.
str = "012za}/n}ddfsdfk"
str[:str.find("}/n}")+4]
Will result in 012za}/n}

Regex issue in python

I have a regex "value=4020a345-f646-4984-a848-3f7f5cb51f21"
if re.search( "value=\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*", x ):
x = re.search( "value=\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*", x )
m = x.group(1)
m only gives me 4020a345, not sure why it does not give me the entire "4020a345-f646-4984-a848-3f7f5cb51f21"
Can anyone tell me what i am doing wrong?

try out this regex, looks like you are trying to match a GUID
value=[0-9a-f]{8}-[0-9a-f]{4}-[0-9a-f]{4}-[0-9a-f]{4}-[0-9a-f]{12}

This should match what you want, if all the strings are of the form you've shown:
value=((\w*\d*\-?)*)
You can also use this website to validate your regular expressions:
http://regex101.com/

The below regex works as you expect.
value=([\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*]+)

You are trying to match on some hex numbers, that is why this regex is more correct than using [\w\d]
pattern = "value=([0-9a-fA-F]{8}-([0-9a-fA-F]{4}-){3}[0-9a-fA-F]{12})"
data = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
res = re.search(pattern, data)
print(res.group(1))
If you dont care about the regex safety, aka checking that it is correct hex, there is no reason not to use simple string manipulation like shown below.
>>> data = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
>>> print(data[7:])
020a345-f646-4984-a848-3f7f5cb51f21
>>> # or maybe
...
>>> print(data[7:].replace('-',''))
020a345f6464984a8483f7f5cb51f21

You can get the subparts of the value as a list
txt = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
parts = re.findall('\w+', txt)[1:]
parts is ['4020a345', 'f646', '4984', 'a848', '3f7f5cb51f21']
if you really want the entire string
full = "-".join(parts)
A simple way
full = re.findall("[\w-]+", txt)[-1]
full is 4020a345-f646-4984-a848-3f7f5cb51f21

value=([\w\d]*\-[\w\d]*\-[\w\d]*\-[\w\d]*\-[\w\d]*)
Try this.Grab the capture.Your regex was not giving the whole as you had used | operator.So if regex on left side of | get satisfied it will not try the latter part.
See demo.
http://regex101.com/r/hQ1rP0/45

python regex for repeating string

I am wanting to verify and then parse this string (in quotes):
string = "start: c12354, c3456, 34526; other stuff that I don't care about"
//Note that some codes begin with 'c'
I would like to verify that the string starts with 'start:' and ends with ';'
Afterward, I would like to have a regex parse out the strings. I tried the following python re code:
regx = r"start: (c?[0-9]+,?)+;"
reg = re.compile(regx)
matched = reg.search(string)
print ' matched.groups()', matched.groups()
I have tried different variations but I can either get the first or the last code but not a list of all three.
Or should I abandon using a regex?
EDIT: updated to reflect part of the problem space I neglected and fixed string difference.
Thanks for all the suggestions - in such a short time.

In Python, this isn’t possible with a single regular expression: each capture of a group overrides the last capture of that same group (in .NET, this would actually be possible since the engine distinguishes between captures and groups).
Your easiest solution is to first extract the part between start: and ; and then using a regular expression to return all matches, not just a single match, using re.findall('c?[0-9]+', text).

You could use the standard string tools, which are pretty much always more readable.
s = "start: c12354, c3456, 34526;"
s.startswith("start:") # returns a boolean if it starts with this string
s.endswith(";") # returns a boolean if it ends with this string
s[6:-1].split(', ') # will give you a list of tokens separated by the string ", "

This can be done (pretty elegantly) with a tool like Pyparsing:
from pyparsing import Group, Literal, Optional, Word
import string
code = Group(Optional(Literal("c"), default='') + Word(string.digits) + Optional(Literal(","), default=''))
parser = Literal("start:") + OneOrMore(code) + Literal(";")
# Read lines from file:
with open('lines.txt', 'r') as f:
for line in f:
try:
result = parser.parseString(line)
codes = [c[1] for c in result[1:-1]]
# Do something with teh codez...
except ParseException exc:
# Oh noes: string doesn't match!
continue
Cleaner than a regular expression, returns a list of codes (no need to string.split), and ignores any extra characters in the line, just like your example.

import re
sstr = re.compile(r'start:([^;]*);')
slst = re.compile(r'(?:c?)(\d+)')
mystr = "start: c12354, c3456, 34526; other stuff that I don't care about"
match = re.match(sstr, mystr)
if match:
res = re.findall(slst, match.group(0))
results in
['12354', '3456', '34526']