Python string multireplacement

multireplace.py

def multireplace(string, replacements, ignore_case=False):
    """
    Given a string and a replacement map, it returns the replaced string.

    :param str string: string to execute replacements on
    :param dict replacements: replacement dictionary {value to find: value to replace}
    :param bool ignore_case: whether the match should be case insensitive
    :rtype: str

    """
    if not replacements:
        # Edge case that'd produce a funny regex and cause a KeyError
        return string
    
    # If case insensitive, we need to normalize the old string so that later a replacement
    # can be found. For instance with {"HEY": "lol"} we should match and find a replacement for "hey",
    # "HEY", "hEy", etc.
    if ignore_case:
        def normalize_old(s):
            return s.lower()

        re_mode = re.IGNORECASE

    else:
        def normalize_old(s):
            return s

        re_mode = 0

    replacements = {normalize_old(key): val for key, val in replacements.items()}
    
    # Place longer ones first to keep shorter substrings from matching where the longer ones should take place
    # For instance given the replacements {'ab': 'AB', 'abc': 'ABC'} against the string 'hey abc', it should produce
    # 'hey ABC' and not 'hey ABc'
    rep_sorted = sorted(replacements, key=len, reverse=True)
    rep_escaped = map(re.escape, rep_sorted)
    
    # Create a big OR regex that matches any of the substrings to replace
    pattern = re.compile("|".join(rep_escaped), re_mode)
    
    # For each match, look up the new string in the replacements, being the key the normalized old string
    return pattern.sub(lambda match: replacements[normalize_old(match.group(0))], string)

Thanks to all who have contributed to this script! One further improvement is to separate the compilation of the regular expression from the processing of the pattern on a string. This provides a big performance improvement when the same regular expression needs to be used to process multiple strings.

Here's my stab at doing this:

import re


def identity(s):
    return s


def lowercase(s):
    return s.lower()


class StringReplacer(object):

    def process(self, string):
        """
        Process the given string by replacing values as configured
        :param str string: string to perform replacements on
        :rtype: str
        """
        replacements = self.replacements
        normalize = self.normalize
        # For each match, look up the new string in the replacements via the normalized old string
        return self.pattern.sub(lambda match: replacements[normalize(match.group(0))], string)

    def __init__(self, replacements, ignore_case=False):
        """
        Given a replacement map, instantiate a StringReplacer.
        :param dict replacements: replacement dictionary {value to find: value to replace}
        :param bool ignore_case: whether the match should be case insensitive
        :rtype: None
        """
        self.normalize = self._configure_normalize(ignore_case)
        self.replacements = self._configure_replacements(replacements, self.normalize)
        self.pattern = self._configure_pattern(self.replacements, ignore_case)

    def _configure_normalize(self, ignore_case):
        # If case insensitive, we need to normalize the old string so that later a replacement
        # can be found. For instance with {"HEY": "lol"} we should match and find a replacement for "hey",
        # "HEY", "hEy", etc.
        return lowercase if ignore_case else identity

    def _configure_replacements(self, replacements, normalize):
        return {normalize(key): value for key, value in replacements.items()}

    def _configure_pattern(self, replacements, ignore_case):
        # Place longer ones first to keep shorter substrings from matching where the longer ones should take place
        # For instance given the replacements {'ab': 'AB', 'abc': 'ABC'} against the string 'hey abc', it should produce
        # 'hey ABC' and not 'hey ABc'
        sorted_replacement_keys = sorted(replacements, key=len, reverse=True)
        escaped_replacement_keys = [re.escape(key) for key in sorted_replacement_keys]

        re_mode = re.IGNORECASE if ignore_case else 0

        # Create a big OR regex that matches any of the substrings to replace
        return re.compile('|'.join(escaped_replacement_keys), re_mode)

@elidchan re.compile caches to a certain extent:

https://docs.python.org/3/library/re.html#re.compile

The compiled versions of the most recent patterns passed to re.compile() and the module-level
matching functions are cached, so programs that use only a few regular expressions at a time
needn’t worry about compiling regular expressions.

So your approach may or may not make sense depending on the scenario (and it could be that you'd have to cache your StringReplacer instances)

Based on @bgusach and @elidchan proposals, I have created a version with support for basic regex replacement. The main restriction is that expressions must not contain subgroups, and there may be some edge cases:

import re

class StringReplacer:

    def __init__(self, replacements, ignore_case=False):
        patterns = sorted(replacements, key=len, reverse=True)
        self.replacements = [replacements[k] for k in patterns]
        re_mode = re.IGNORECASE if ignore_case else 0
        self.pattern = re.compile('|'.join(("({})".format(p) for p in patterns)), re_mode)
        def tr(matcher):
            index = next((index for index,value in enumerate(matcher.groups()) if value), None)
            return self.replacements[index]
        self.tr = tr

    def __call__(self, string):
        return self.pattern.sub(self.tr, string)

Tests

table = {
    "aaa"    : "[This is three a]",
    "b+"     : "[This is one or more b]",
    r"<\w+>" : "[This is a tag]"
}

replacer = StringReplacer(table, True)

sample1 = "whatever bb, aaa, <star> BBB <end>"

print(replacer(sample1))

# output: whatever [This is one or more b], [This is three a], [This is a tag] [This is one or more b] [This is a tag]

The trick is to identify the matched group by its position. It is not super efficient (O(n)), but it works.

index = next((index for index,value in enumerate(matcher.groups()) if value), None)

Replacement is done in one pass.

How would one apply multireplace to strings in pandas dataframe?

Based on @mnesarco approach, I tried a functional one with support for one subgroup per expression:

import re
from typing import Dict, Union


def multireplace(table: Dict[str, str], string: str, flags: Union[int, re.RegexFlag] = 0):
    patterns = {
        f"_g{n}": pattern for n, pattern in enumerate(table)
    }

    def repl(match: re.Match):
        repkey = None
        groupkey = None

        for key, value in match.groupdict().items():
            if value is None:
                continue

            if key.startswith("_g"):
                repkey = key
            else:
                groupkey, groupval = key, value

        repval = table[patterns[repkey]]
        return repval if groupkey is None else repval.replace(fr"\g<{groupkey}>", groupval)

    regex = "|".join(fr"(?P<{group}>{rep})" for group, rep in patterns.items())
    return re.sub(regex, repl, string, flags=flags)

Test

table = {
    "aaa": "[This is three a]",
    "b+": "[This is one or more b]",
    r"(?<=<spam>).+(?=</spam>)": "[REDACTED]",
    r"</?\w+>": "[This is a tag]",
}

txt = multireplace(table, "whatever bb, aaa, <star> BBB <end> <tag>keep me</tag> and <spam>delete me</spam>", re.IGNORECASE)
print(txt)
# output: whatever [This is one or more b], [This is three a], [This is a tag] [This is one or more b] [This is a tag] [This is a tag]keep me[This is a tag] and [This is a tag][REDACTED][This is a tag]

table = {
    "aaa": "[This is three a]",
    "b+": "[This is one or more b]",
    r"<(?P<name>\w+)>(?P<value>.+)</(?P=name)>": r"[This is an HTML tag with text (\g<value>)]",
    r"</?\w+>": "[This is a tag]",
}

txt = multireplace(table, "whatever bb, aaa, <star> BBB <end> <tag>keep me</tag> and <spam>delete me</spam>", re.IGNORECASE)
print(txt)

# output: whatever [This is one or more b], [This is three a], [This is a tag] [This is one or more b] [This is a tag] [This is an HTML tag with text (keep me)] and [This is an HTML tag with text (delete me)]

It's still O(n), I don't know how priorities are being set inside the main regex, they should be based on the dictionary order, but when there is competition (eg r"<(?P<name>\w+)>(?P<value>.+)</(?P=name)>" versus r"(?<=<spam>).+(?=</spam>)") the first has precedence. Also, one cannot reference a group by its order, only by name.