Speed up millions of regex replacements in Python 3

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Speed up millions of regex replacements in Python 3

I have two lists:

  • a list of about 750K “sentences” (long strings)
  • a list of about 20K “words” that I would like to delete from my 750K sentences

So, I have to loop through 750K sentences and perform about 20K replacements, but ONLY if my words are actually “words” and are not part of a larger string of characters.

I am doing this by pre-compiling my words so that they are flanked by the b word-boundary metacharacter:

compiled_words = [re.compile(r'b' + word + r'b') for word in my20000words]

Then I loop through my “sentences”:

import re
for sentence in sentences:
  for word in compiled_words:
    sentence = re.sub(word, "", sentence)
  # put sentence into a growing list

This nested loop is processing about 50 sentences per second, which is nice, but it still takes several hours to process all of my sentences.

  • Is there a way to using the str.replace method (which I believe is faster), but still requiring that replacements only happen at word boundaries?

  • Alternatively, is there a way to speed up the re.sub method? I have already improved the speed marginally by skipping over re.sub if the length of my word is > than the length of my sentence, but it’s not much of an improvement.

I’m using Python 3.5.2

Asked By: pdanese

||

Answer #1:

One thing you can try is to compile one single pattern like "b(word1|word2|word3)b".

Because re relies on C code to do the actual matching, the savings can be dramatic.

As @pvg pointed out in the comments, it also benefits from single pass matching.

If your words are not regex, Eric’s answer is faster.

Answered By: Liteye

Answer #2:

TLDR

Use this method (with set lookup) if you want the fastest solution. For a dataset similar to the OP’s, it’s approximately 2000 times faster than the accepted answer.

If you insist on using a regex for lookup, use this trie-based version, which is still 1000 times faster than a regex union.

Theory

If your sentences aren’t humongous strings, it’s probably feasible to process many more than 50 per second.

If you save all the banned words into a set, it will be very fast to check if another word is included in that set.

Pack the logic into a function, give this function as argument to re.sub and you’re done!

Code

import re
with open('/usr/share/dict/american-english') as wordbook:
    banned_words = set(word.strip().lower() for word in wordbook)
def delete_banned_words(matchobj):
    word = matchobj.group(0)
    if word.lower() in banned_words:
        return ""
    else:
        return word
sentences = ["I'm eric. Welcome here!", "Another boring sentence.",
             "GiraffeElephantBoat", "sfgsdg sdwerha aswertwe"] * 250000
word_pattern = re.compile('w+')
for sentence in sentences:
    sentence = word_pattern.sub(delete_banned_words, sentence)

Converted sentences are:

' .  !
  .
GiraffeElephantBoat
sfgsdg sdwerha aswertwe

Note that:

  • the search is case-insensitive (thanks to lower())
  • replacing a word with "" might leave two spaces (as in your code)
  • With python3, w+ also matches accented characters (e.g. "ångström").
  • Any non-word character (tab, space, newline, marks, …) will stay untouched.

Performance

There are a million sentences, banned_words has almost 100000 words and the script runs in less than 7s.

In comparison, Liteye’s answer needed 160s for 10 thousand sentences.

With n being the total amound of words and m the amount of banned words, OP’s and Liteye’s code are O(n*m).

In comparison, my code should run in O(n+m). Considering that there are many more sentences than banned words, the algorithm becomes O(n).

Regex union test

What’s the complexity of a regex search with a 'b(word1|word2|...|wordN)b' pattern? Is it O(N) or O(1)?

It’s pretty hard to grasp the way the regex engine works, so let’s write a simple test.

This code extracts 10**i random english words into a list. It creates the corresponding regex union, and tests it with different words :

  • one is clearly not a word (it begins with #)
  • one is the first word in the list
  • one is the last word in the list
  • one looks like a word but isn’t

import re
import timeit
import random
with open('/usr/share/dict/american-english') as wordbook:
    english_words = [word.strip().lower() for word in wordbook]
    random.shuffle(english_words)
print("First 10 words :")
print(english_words[:10])
test_words = [
    ("Surely not a word", "#surely_NöTäWORD_so_regex_engine_can_return_fast"),
    ("First word", english_words[0]),
    ("Last word", english_words[-1]),
    ("Almost a word", "couldbeaword")
]
def find(word):
    def fun():
        return union.match(word)
    return fun
for exp in range(1, 6):
    print("nUnion of %d words" % 10**exp)
    union = re.compile(r"b(%s)b" % '|'.join(english_words[:10**exp]))
    for description, test_word in test_words:
        time = timeit.timeit(find(test_word), number=1000) * 1000
        print("  %-17s : %.1fms" % (description, time))

It outputs:

First 10 words :
["geritol's", "sunstroke's", 'fib', 'fergus', 'charms', 'canning', 'supervisor', 'fallaciously', "heritage's", 'pastime']
Union of 10 words
  Surely not a word : 0.7ms
  First word        : 0.8ms
  Last word         : 0.7ms
  Almost a word     : 0.7ms
Union of 100 words
  Surely not a word : 0.7ms
  First word        : 1.1ms
  Last word         : 1.2ms
  Almost a word     : 1.2ms
Union of 1000 words
  Surely not a word : 0.7ms
  First word        : 0.8ms
  Last word         : 9.6ms
  Almost a word     : 10.1ms
Union of 10000 words
  Surely not a word : 1.4ms
  First word        : 1.8ms
  Last word         : 96.3ms
  Almost a word     : 116.6ms
Union of 100000 words
  Surely not a word : 0.7ms
  First word        : 0.8ms
  Last word         : 1227.1ms
  Almost a word     : 1404.1ms

So it looks like the search for a single word with a 'b(word1|word2|...|wordN)b' pattern has:

  • O(1) best case
  • O(n/2) average case, which is still O(n)
  • O(n) worst case

These results are consistent with a simple loop search.

A much faster alternative to a regex union is to create the regex pattern from a trie.

Answered By: Eric Duminil

Answer #3:

TLDR

Use this method if you want the fastest regex-based solution. For a dataset similar to the OP’s, it’s approximately 1000 times faster than the accepted answer.

If you don’t care about regex, use this set-based version, which is 2000 times faster than a regex union.

Optimized Regex with Trie

A simple Regex union approach becomes slow with many banned words, because the regex engine doesn’t do a very good job of optimizing the pattern.

It’s possible to create a Trie with all the banned words and write the corresponding regex. The resulting trie or regex aren’t really human-readable, but they do allow for very fast lookup and match.

Example

['foobar', 'foobah', 'fooxar', 'foozap', 'fooza']

Regex union

The list is converted to a trie:

{
    'f': {
        'o': {
            'o': {
                'x': {
                    'a': {
                        'r': {
                            '': 1
                        }
                    }
                },
                'b': {
                    'a': {
                        'r': {
                            '': 1
                        },
                        'h': {
                            '': 1
                        }
                    }
                },
                'z': {
                    'a': {
                        '': 1,
                        'p': {
                            '': 1
                        }
                    }
                }
            }
        }
    }
}

And then to this regex pattern:

r"bfoo(?:ba[hr]|xar|zap?)b"

Regex trie

The huge advantage is that to test if zoo matches, the regex engine only needs to compare the first character (it doesn’t match), instead of trying the 5 words. It’s a preprocess overkill for 5 words, but it shows promising results for many thousand words.

Note that (?:) non-capturing groups are used because:

Code

Here’s a slightly modified gist, which we can use as a trie.py library:

import re
class Trie():
    """Regex::Trie in Python. Creates a Trie out of a list of words. The trie can be exported to a Regex pattern.
    The corresponding Regex should match much faster than a simple Regex union."""
    def __init__(self):
        self.data = {}
    def add(self, word):
        ref = self.data
        for char in word:
            ref[char] = char in ref and ref[char] or {}
            ref = ref[char]
        ref[''] = 1
    def dump(self):
        return self.data
    def quote(self, char):
        return re.escape(char)
    def _pattern(self, pData):
        data = pData
        if "" in data and len(data.keys()) == 1:
            return None
        alt = []
        cc = []
        q = 0
        for char in sorted(data.keys()):
            if isinstance(data[char], dict):
                try:
                    recurse = self._pattern(data[char])
                    alt.append(self.quote(char) + recurse)
                except:
                    cc.append(self.quote(char))
            else:
                q = 1
        cconly = not len(alt) > 0
        if len(cc) > 0:
            if len(cc) == 1:
                alt.append(cc[0])
            else:
                alt.append('[' + ''.join(cc) + ']')
        if len(alt) == 1:
            result = alt[0]
        else:
            result = "(?:" + "|".join(alt) + ")"
        if q:
            if cconly:
                result += "?"
            else:
                result = "(?:%s)?" % result
        return result
    def pattern(self):
        return self._pattern(self.dump())

Test

Here’s a small test (the same as this one):

# Encoding: utf-8
import re
import timeit
import random
from trie import Trie
with open('/usr/share/dict/american-english') as wordbook:
    banned_words = [word.strip().lower() for word in wordbook]
    random.shuffle(banned_words)
test_words = [
    ("Surely not a word", "#surely_NöTäWORD_so_regex_engine_can_return_fast"),
    ("First word", banned_words[0]),
    ("Last word", banned_words[-1]),
    ("Almost a word", "couldbeaword")
]
def trie_regex_from_words(words):
    trie = Trie()
    for word in words:
        trie.add(word)
    return re.compile(r"b" + trie.pattern() + r"b", re.IGNORECASE)
def find(word):
    def fun():
        return union.match(word)
    return fun
for exp in range(1, 6):
    print("nTrieRegex of %d words" % 10**exp)
    union = trie_regex_from_words(banned_words[:10**exp])
    for description, test_word in test_words:
        time = timeit.timeit(find(test_word), number=1000) * 1000
        print("  %s : %.1fms" % (description, time))

It outputs:

TrieRegex of 10 words
  Surely not a word : 0.3ms
  First word : 0.4ms
  Last word : 0.5ms
  Almost a word : 0.5ms
TrieRegex of 100 words
  Surely not a word : 0.3ms
  First word : 0.5ms
  Last word : 0.9ms
  Almost a word : 0.6ms
TrieRegex of 1000 words
  Surely not a word : 0.3ms
  First word : 0.7ms
  Last word : 0.9ms
  Almost a word : 1.1ms
TrieRegex of 10000 words
  Surely not a word : 0.1ms
  First word : 1.0ms
  Last word : 1.2ms
  Almost a word : 1.2ms
TrieRegex of 100000 words
  Surely not a word : 0.3ms
  First word : 1.2ms
  Last word : 0.9ms
  Almost a word : 1.6ms

For info, the regex begins like this:

(?:a(?:(?:’s|a(?:’s|chen|liyah(?:’s)?|r(?:dvark(?:(?:’s|s))?|on))|b(?:’s|a(?:c(?:us(?:(?:’s|es))?|[ik])|ft|lone(?:(?:’s|s))?|ndon(?:(?:ed|ing|ment(?:’s)?|s))?|s(?:e(?:(?:ment(?:’s)?|[ds]))?|h(?:(?:e[ds]|ing))?|ing)|t(?:e(?:(?:ment(?:’s)?|[ds]))?|ing|toir(?:(?:’s|s))?))|b(?:as(?:id)?|e(?:ss(?:(?:’s|es))?|y(?:(?:’s|s))?)|ot(?:(?:’s|t(?:’s)?|s))?|reviat(?:e[ds]?|i(?:ng|on(?:(?:’s|s))?))|y(?:’s)?|é(?:(?:’s|s))?)|d(?:icat(?:e[ds]?|i(?:ng|on(?:(?:’s|s))?))|om(?:en(?:(?:’s|s))?|inal)|u(?:ct(?:(?:ed|i(?:ng|on(?:(?:’s|s))?)|or(?:(?:’s|s))?|s))?|l(?:’s)?))|e(?:(?:’s|am|l(?:(?:’s|ard|son(?:’s)?))?|r(?:deen(?:’s)?|nathy(?:’s)?|ra(?:nt|tion(?:(?:’s|s))?))|t(?:(?:t(?:e(?:r(?:(?:’s|s))?|d)|ing|or(?:(?:’s|s))?)|s))?|yance(?:’s)?|d))?|hor(?:(?:r(?:e(?:n(?:ce(?:’s)?|t)|d)|ing)|s))?|i(?:d(?:e[ds]?|ing|jan(?:’s)?)|gail|l(?:ene|it(?:ies|y(?:’s)?)))|j(?:ect(?:ly)?|ur(?:ation(?:(?:’s|s))?|e[ds]?|ing))|l(?:a(?:tive(?:(?:’s|s))?|ze)|e(?:(?:st|r))?|oom|ution(?:(?:’s|s))?|y)|m’s|n(?:e(?:gat(?:e[ds]?|i(?:ng|on(?:’s)?))|r(?:’s)?)|ormal(?:(?:it(?:ies|y(?:’s)?)|ly))?)|o(?:ard|de(?:(?:’s|s))?|li(?:sh(?:(?:e[ds]|ing))?|tion(?:(?:’s|ist(?:(?:’s|s))?))?)|mina(?:bl[ey]|t(?:e[ds]?|i(?:ng|on(?:(?:’s|s))?)))|r(?:igin(?:al(?:(?:’s|s))?|e(?:(?:’s|s))?)|t(?:(?:ed|i(?:ng|on(?:(?:’s|ist(?:(?:’s|s))?|s))?|ve)|s))?)|u(?:nd(?:(?:ed|ing|s))?|t)|ve(?:(?:’s|board))?)|r(?:a(?:cadabra(?:’s)?|d(?:e[ds]?|ing)|ham(?:’s)?|m(?:(?:’s|s))?|si(?:on(?:(?:’s|s))?|ve(?:(?:’s|ly|ness(?:’s)?|s))?))|east|idg(?:e(?:(?:ment(?:(?:’s|s))?|[ds]))?|ing|ment(?:(?:’s|s))?)|o(?:ad|gat(?:e[ds]?|i(?:ng|on(?:(?:’s|s))?)))|upt(?:(?:e(?:st|r)|ly|ness(?:’s)?))?)|s(?:alom|c(?:ess(?:(?:’s|e[ds]|ing))?|issa(?:(?:’s|[es]))?|ond(?:(?:ed|ing|s))?)|en(?:ce(?:(?:’s|s))?|t(?:(?:e(?:e(?:(?:’s|ism(?:’s)?|s))?|d)|ing|ly|s))?)|inth(?:(?:’s|e(?:’s)?))?|o(?:l(?:ut(?:e(?:(?:’s|ly|st?))?|i(?:on(?:’s)?|sm(?:’s)?))|v(?:e[ds]?|ing))|r(?:b(?:(?:e(?:n(?:cy(?:’s)?|t(?:(?:’s|s))?)|d)|ing|s))?|pti…

It’s really unreadable, but for a list of 100000 banned words, this Trie regex is 1000 times faster than a simple regex union!

Here’s a diagram of the complete trie, exported with trie-python-graphviz and graphviz twopi:

Enter image description here

Answered By: Eric Duminil

Answer #4:

One thing you might want to try is pre-processing the sentences to encode the word boundaries. Basically turn each sentence into a list of words by splitting on word boundaries.

This should be faster, because to process a sentence, you just have to step through each of the words and check if it’s a match.

Currently the regex search is having to go through the entire string again each time, looking for word boundaries, and then “discarding” the result of this work before the next pass.

Answered By: Denziloe

Answer #5:

Well, here’s a quick and easy solution, with test set.

Best strategy:

  • re.sub("w+",repl,sentence) searches for words.
  • “repl” can be a callable. I used a function that performs a dict lookup, and the dict contains the words to search and replace.
  • This is the simplest and fastest solution (see function replace4 in example code below).

Second best strategy:

  • The idea is to split the sentences into words, using re.split, while conserving the separators to reconstruct the sentences later. Then, replacements are done with a simple dict lookup.
  • Implementation: (see function replace3 in example code below).

Timings for example functions:

replace1:     0.62 sentences/s
replace2:     7.43 sentences/s
replace3: 48498.03 sentences/s
replace4: 61374.97 sentences/s (...and 240,000/s with PyPy)

…and code:

#! /bin/env python3
# -*- coding: utf-8
import time, random, re
def replace1( sentences ):
    for n, sentence in enumerate( sentences ):
        for search, repl in patterns:
            sentence = re.sub( "\b"+search+"\b", repl, sentence )
def replace2( sentences ):
    for n, sentence in enumerate( sentences ):
        for search, repl in patterns_comp:
            sentence = re.sub( search, repl, sentence )
def replace3( sentences ):
    pd = patterns_dict.get
    for n, sentence in enumerate( sentences ):
        #~ print( n, sentence )
        # Split the sentence on non-word characters.
        # Note: () in split patterns ensure the non-word characters ARE kept
        # and returned in the result list, so we don't mangle the sentence.
        # If ALL separators are spaces, use string.split instead or something.
        # Example:
        #~ >>> re.split(r"([^w]+)", "ab céé? . d2eéf")
        #~ ['ab', ' ', 'céé', '? . ', 'd2eéf']
        words = re.split(r"([^w]+)", sentence)
        # and... done.
        sentence = "".join( pd(w,w) for w in words )
        #~ print( n, sentence )
def replace4( sentences ):
    pd = patterns_dict.get
    def repl(m):
        w = m.group()
        return pd(w,w)
    for n, sentence in enumerate( sentences ):
        sentence = re.sub(r"w+", repl, sentence)
# Build test set
test_words = [ ("word%d" % _) for _ in range(50000) ]
test_sentences = [ " ".join( random.sample( test_words, 10 )) for _ in range(1000) ]
# Create search and replace patterns
patterns = [ (("word%d" % _), ("repl%d" % _)) for _ in range(20000) ]
patterns_dict = dict( patterns )
patterns_comp = [ (re.compile("\b"+search+"\b"), repl) for search, repl in patterns ]
def test( func, num ):
    t = time.time()
    func( test_sentences[:num] )
    print( "%30s: %.02f sentences/s" % (func.__name__, num/(time.time()-t)))
print( "Sentences", len(test_sentences) )
print( "Words    ", len(test_words) )
test( replace1, 1 )
test( replace2, 10 )
test( replace3, 1000 )
test( replace4, 1000 )

EDIT: You can also ignore lowercase when checking if you pass a lowercase list of Sentences and edit repl

def replace4( sentences ):
pd = patterns_dict.get
def repl(m):
    w = m.group()
    return pd(w.lower(),w)
Answered By: bobflux

Answer #6:

Perhaps Python is not the right tool here. Here is one with the Unix toolchain

sed G file         |
tr ' ' 'n'        |
grep -vf blacklist |
awk -v RS= -v OFS=' ' '{$1=$1}1'

assuming your blacklist file is preprocessed with the word boundaries added. The steps are: convert the file to double spaced, split each sentence to one word per line, mass delete the blacklist words from the file, and merge back the lines.

This should run at least an order of magnitude faster.

For preprocessing the blacklist file from words (one word per line)

sed 's/.*/\b&\b/' words > blacklist
Answered By: karakfa

Answer #7:

How about this:

#!/usr/bin/env python3
from __future__ import unicode_literals, print_function
import re
import time
import io
def replace_sentences_1(sentences, banned_words):
    # faster on CPython, but does not use b as the word separator
    # so result is slightly different than replace_sentences_2()
    def filter_sentence(sentence):
        words = WORD_SPLITTER.split(sentence)
        words_iter = iter(words)
        for word in words_iter:
            norm_word = word.lower()
            if norm_word not in banned_words:
                yield word
            yield next(words_iter) # yield the word separator
    WORD_SPLITTER = re.compile(r'(W+)')
    banned_words = set(banned_words)
    for sentence in sentences:
        yield ''.join(filter_sentence(sentence))
def replace_sentences_2(sentences, banned_words):
    # slower on CPython, uses b as separator
    def filter_sentence(sentence):
        boundaries = WORD_BOUNDARY.finditer(sentence)
        current_boundary = 0
        while True:
            last_word_boundary, current_boundary = current_boundary, next(boundaries).start()
            yield sentence[last_word_boundary:current_boundary] # yield the separators
            last_word_boundary, current_boundary = current_boundary, next(boundaries).start()
            word = sentence[last_word_boundary:current_boundary]
            norm_word = word.lower()
            if norm_word not in banned_words:
                yield word
    WORD_BOUNDARY = re.compile(r'b')
    banned_words = set(banned_words)
    for sentence in sentences:
        yield ''.join(filter_sentence(sentence))
corpus = io.open('corpus2.txt').read()
banned_words = [l.lower() for l in open('banned_words.txt').read().splitlines()]
sentences = corpus.split('. ')
output = io.open('output.txt', 'wb')
print('number of sentences:', len(sentences))
start = time.time()
for sentence in replace_sentences_1(sentences, banned_words):
    output.write(sentence.encode('utf-8'))
    output.write(b' .')
print('time:', time.time() - start)

These solutions splits on word boundaries and looks up each word in a set. They should be faster than re.sub of word alternates (Liteyes’ solution) as these solutions are O(n) where n is the size of the input due to the amortized O(1) set lookup, while using regex alternates would cause the regex engine to have to check for word matches on every characters rather than just on word boundaries. My solutiona take extra care to preserve the whitespaces that was used in the original text (i.e. it doesn’t compress whitespaces and preserves tabs, newlines, and other whitespace characters), but if you decide that you don’t care about it, it should be fairly straightforward to remove them from the output.

I tested on corpus.txt, which is a concatenation of multiple eBooks downloaded from the Gutenberg Project, and banned_words.txt is 20000 words randomly picked from Ubuntu’s wordlist (/usr/share/dict/american-english). It takes around 30 seconds to process 862462 sentences (and half of that on PyPy). I’ve defined sentences as anything separated by “. “.

$ # replace_sentences_1()
$ python3 filter_words.py
number of sentences: 862462
time: 24.46173644065857
$ pypy filter_words.py
number of sentences: 862462
time: 15.9370770454
$ # replace_sentences_2()
$ python3 filter_words.py
number of sentences: 862462
time: 40.2742919921875
$ pypy filter_words.py
number of sentences: 862462
time: 13.1190629005

PyPy particularly benefit more from the second approach, while CPython fared better on the first approach. The above code should work on both Python 2 and 3.

Answered By: Lie Ryan

Answer #8:

Practical approach

A solution described below uses a lot of memory to store all the text at the same string and to reduce complexity level. If RAM is an issue think twice before use it.

With join/split tricks you can avoid loops at all which should speed up the algorithm.

  • Concatenate a sentences with a special delimeter which is not contained by the sentences:
  • merged_sentences = ' * '.join(sentences)
    

  • Compile a single regex for all the words you need to rid from the sentences using | “or” regex statement:
  • regex = re.compile(r'b({})b'.format('|'.join(words)), re.I) # re.I is a case insensitive flag
    

  • Subscript the words with the compiled regex and split it by the special delimiter character back to separated sentences:
  • clean_sentences = re.sub(regex, "", merged_sentences).split(' * ')
    

    Performance

    "".join complexity is O(n). This is pretty intuitive but anyway there is a shortened quotation from a source:

    for (i = 0; i < seqlen; i++) {
        [...]
        sz += PyUnicode_GET_LENGTH(item);
    

    Therefore with join/split you have O(words) + 2*O(sentences) which is still linear complexity vs 2*O(N2) with the initial approach.


    b.t.w. don’t use multithreading. GIL will block each operation because your task is strictly CPU bound so GIL have no chance to be released but each thread will send ticks concurrently which cause extra effort and even lead operation to infinity.

    Answered By: I159

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