Replace backslash followed by double quotation in a text file in Python

I have a text file, and its content is like this:

"good to know it \" so nice \" "

I use Python to read its contents and want to replace " with an empty string.

The code I am using is:

import re

file_path = "backslash_double_quotation.txt"
with open(file_path, "r") as input_file:
    raw_text = input_file.read()
processed_text = re.sub(r'\"', "", raw_text)
print(raw_text)
print(processed_text)

and I expect processed_text like this:

"good to know it  so nice  "

However, the actual output is:

good to know it \ so nice \

All the double quotations are replaced by empty strings.
How can I fix this?

With strings you can use .replace() to replace specific characters or words in a string.

For example:

text = "good to know it \" so nice \""
print(text.replace("\"", " "))

The output for this is:

good to know it   so nice  

With your code:

import re
file_path = "backslash_double_quotation.txt"
with open(file_path, "r") as input_file:
    raw_text = input_file.read()
processed_text = raw_text.replace("\"", "")
print(raw_text)
print(processed_text)

If you want to use re then:

processed_text = re.sub(r"\\", "", raw_text)

You don't get the expected result because of "raw-string", "r" in your example. If you add "r" you should specify your regex expression without any escape characters.

Just remove "r" in your example and it will work as expected:

processed_text = re.sub('\"', "", raw_text)

Reference:

Raw String Notation

Eliminate one by one

processed_text = raw_text.replace('"', '')
processed_text = processed_text.replace('\', '')

It's hard to imagine that an escaped double quote \" means something else than include this quote in the delimited double quote string. Therefore it's impossible to imagine not using an escaped escape \\ to differentiate an included escape in the string from not treating a following double quote (if any) as the closing string delimiter.

This seems to be a nonambiguous way to tell the difference -

https://regex101.com/r/FH2Dfp/1

Find (raw context, wrap in r' '):

(?<!\\)((?:\\\\)*)\\"

Replace with:

\1

I found this works:

processed_text = re.sub(r'\\"', "", raw_text)