如何去除不需要的字符和字符串？

Question

I want to strip all unwanted [AZ] characters (among others) except for certain words. For example, we have the following string:

get 5 and 9

I would like to get rid of all the words that are not 'and' or 'or' so the end result is 5 and 9 . I also want to strip out all characters not being part of '[0-9].+-*()<>\\s' too.

The current regular expression works for stripping out all characters but I don't want it to strip out 'and'. In this example, the result would be '5 9'.

string = 'get 5 and 9'
pattern = re.compile(r'[^0-9\.\+\-\/\*\(\)<>\s)]')
string = re.sub(pattern, '', string)

I am not an expert on regular expressions and struggle to find a solution for this. I am kind of lost.

Is this possible or should I look for other solutions?

Answer 1

Revised version

import re

test = "get 6 AND 9 or 3 for 6"
keywords = ['and', 'or']
print(' '.join(t for t in test.split() if t.lower() in keywords or t.isdigit()))

$ python test.py
6 AND 9 or 3 6

This rejects words containing and and or,

Previous version. This is a pretty simple solution I think, but unfortunately did not work as it picks up 'and' and 'or' in longer words.

import re

test = "get 6 AND 9 or 3"
pattern=re.compile("(?i)(and|or|\d|\s)")
result = re.findall(pattern, test)
print(''.join(result).strip())

$ python test.py
6 AND 9 or 3

Words are case-insensitive because of (?i). Spaces are retained with \\s but stripped from beginning and end in the print statement. Digits are retained through \\d. The parentheses around and|or|\\d|\\s are the bits of the string that are found through findall which generates a list of what has been found, then they are joined back together in the print function.

Answer 2

An approach without using regular expression

input = 'get 5 and 9'

accept_list = ['and', 'or']

output = []
for x in input.split():
    try :
        output.append(str(int(x)))
    except :
        if x in accept_list:
            output.append(x)

print (' '.join(output))

Output

5 and 9