How to delete brackets and text within them python

Question

I am generating a text file the contents looks like this

[AnyTextOrNumber]AnyTextOrNumber#AnyNumber

How can I generate the file so it always looks like this

AnyTextOrNumber#AnyNumber

I just want to get the text file generated without the brackets and contents within them

Stack Overflow is not a code writing service. Actually try to write some code, and we might be able to help troubleshoot issues you have with it. — Makonede, Jun 27 '22 at 03:41
The code I have gets text from an image in the format above and makes a text file with the generated code. Sometimes the code has leading brackets with letters or numbers in them I am asking for help on how to process the text in the text file to not have the leading Brackets and their contents.... so if someone can explain to me how to do that with sub() id appreciate it. — Billy FReally, Jun 27 '22 at 04:04
I would recommend to have a look at https://docs.python.org/3/library/re.html there are many tutorials on regular expressions around. — Carlos Horn, Jun 27 '22 at 05:51
@BillyFReally You can mark the answer of your choice as accepted if they helped you with your query. — saurabh dhyani, Jun 30 '22 at 08:34

score 1 · Answer 1 · answered Jun 27 '22 at 06:26

You can use re to achieve the desired result.

I've used two patterns and different string scenarios and their behavior. You can see the limitations we can have and chose whichever suits your purpose the best.
You could use something like this :

import re

str1 = "[AnyTextOrNumber]AnyTextOrNumber#AnyNumber"
str2 = "[AnyTextOrNumber]AnyTextOrNumber#AnyNumber[MoreTextOrNumber]"
str3 = "[AnyTextOr]Number]AnyTextOrNumber#AnyNumber"
str4 = "[AnyTextOr]Number]AnyTextOrNumber#AnyNumber[MoreTextOrNumber]"

pattern1 = r'\[[^()]*\]'
pattern2 = r'\[.*?\]'

convPattern1_str1 = re.sub(pattern1, '', str1)
convPattern1_str2 = re.sub(pattern1, '', str2)
convPattern1_str3 = re.sub(pattern1, '', str3)
convPattern1_str4 = re.sub(pattern1, '', str4)

convPattern2_str1 = re.sub(pattern2, '', str1)
convPattern2_str2 = re.sub(pattern2, '', str2)
convPattern2_str3 = re.sub(pattern2, '', str3)
convPattern2_str4 = re.sub(pattern2, '', str4)

print(convPattern1_str1) #Will give AnyTextOrNumber#AnyNumber
print(convPattern1_str2) #Will remove everything
print(convPattern1_str3) #Will give AnyTextOrNumber#AnyNumber
print(convPattern1_str4) #Will remove everything

print(convPattern2_str1) #Will give AnyTextOrNumber#AnyNumber
print(convPattern2_str2) #Will give AnyTextOrNumber#AnyNumber
print(convPattern2_str3) #Will give Number]AnyTextOrNumber#AnyNumber
print(convPattern2_str4) ##Will give Number]AnyTextOrNumber#AnyNumber

score 0 · Answer 2 · answered Jun 27 '22 at 06:22

Here's how you could do it with slicing:

input_file = "file_input.txt"
output_file = "file_output.txt"

# Open 2 files, 1 for reading, 1 for writing
with open(input_file, "r") as in_file, open(output_file, "w") as out_file:
    for row in in_file:
        # Remove newlines
        row = row.strip()
        # find "#"
        number_index = row.find("#")
        # slice from string
        any_number = row[number_index:]
        # find "]" and slice
        string_index = row.find("]")
        any_string = row[1:string_index]
        # Format new line and write to file
        out_file.write(f"{any_string}{any_number}\n")

How to delete brackets and text within them python

2 Answers2