I've been looking for a way to isolate special characters in a regex expression, but I only seem to find the exact opposite of what I'm looking for. So basically I want to is something along the lines of this:
import re
str = "I only want characters from the pattern below to appear in a list ()[]' including quotations"
pattern = """(){}[]"'-"""
result = re.findall(pattern, str)
What I expect from this is:
print(result)
#["(", ")", "[", "]", "'"]
Why would you need regex for this when it can be done without regex?
>>> str = "I only want characters from the pattern below to appear in a list ()[]' including quotations"
>>> pattern = """(){}[]"'-"""
>>> [x for x in str if x in pattern]
['(', ')', '[', ']', "'"]
If it's for learning purposes (regex isn't really the best way here), then you can use:
import re
text = "I only want characters from the pattern below to appear in a list ()[]' including quotations"
output = re.findall('[' + re.escape("""(){}[]"'-""") + ']', text)
# ['(', ')', '[', ']', "'"]
Surrounding the characters in [ and ] makes it a regex character class and re.escape will escape any characters that have special regex meaning to avoid breaking the regex string (eg: ] terminating the characters early or - in a certain place causing it to act like a character range).
Several of the characters in your set have special meaning in regular expressions; to match them literally, you need to backslash-escape them.
pattern = r"""\(\)\{\}\[]"'-"""
Alternatively, you could use a character class:
pattern = """[]-[(){}"']"""
Notice also the use of a "raw string" r'...' to avoid having Python interpret the backslashes.
This regex that solved my problem:
pattern = r"""[(){}\[\]"'\-]"""
Posted on behalf of the question asker
