在Python 3中加速数百万个正则表达式的替换-Python 实用宝典

在Python 3中加速数百万个正则表达式的替换

我正在使用Python 3.5.2 我有两个清单 大约750,000个“句子”(长字符串)的列表 我想从我的750,000个句子中删除的大约20,000个“单词”的列表 因此,我必须遍历750,000个句子并执行大约20,000个替换,但前提是我的单词实际上是“单词”,并且不属于较大的字符串。 我这样做是通过预编译我的单词,使它们位于\b元字符的侧面 compiled_words = [re.compile(r'\b' + word + r'\b') for word in my20000words] 然后我遍历我的“句子” import re for sentence in sentences: for word in compiled_words: sentence = re.sub(word, "", sentence) # put sentence into a growing list 这个嵌套循环每秒处理大约50个句子,这很好,但是处理我所有的句子仍需要几个小时。 有没有一种方法可以使用该str.replace方法(我认为该方法更快),但仍然要求仅在单词边界处进行替换? 或者,有没有办法加快该re.sub方法?re.sub如果单词的长度大于句子的长度,我已经略微提高了速度,但这并没有太大的改进。 感谢您的任何建议。

问题:在Python 3中加速数百万个正则表达式的替换

我正在使用Python 3.5.2

我有两个清单

  • 大约750,000个“句子”(长字符串)的列表
  • 我想从我的750,000个句子中删除的大约20,000个“单词”的列表

因此,我必须遍历750,000个句子并执行大约20,000个替换,但前提是我的单词实际上是“单词”,并且不属于较大的字符串。

我这样做是通过预编译我的单词,使它们位于\b元字符的侧面

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

然后我遍历我的“句子”

import re

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

这个嵌套循环每秒处理大约50个句子,这很好,但是处理我所有的句子仍需要几个小时。

  • 有没有一种方法可以使用该str.replace方法(我认为该方法更快),但仍然要求仅在单词边界处进行替换?

  • 或者,有没有办法加快该re.sub方法?re.sub如果单词的长度大于句子的长度,我已经略微提高了速度,但这并没有太大的改进。

感谢您的任何建议。

I'm using Python 3.5.2

I have two lists

  • a list of about 750,000 "sentences" (long strings)
  • a list of about 20,000 "words" that I would like to delete from my 750,000 sentences

So, I have to loop through 750,000 sentences and perform about 20,000 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 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.

Thank you for any suggestions.


回答 0

您可以尝试做的一件事是编译一个单一模式,例如"\b(word1|word2|word3)\b"

由于re依靠C代码进行实际匹配,因此节省的费用可观。

正如@pvg在评论中指出的,它也受益于单遍匹配。

如果您的单词不是正则表达式,那么Eric的答案会更快。

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.


回答 1

TLDR

如果您想要最快的解决方案,请使用此方法(带有设置的查找)。对于类似于OP的数据集,它比接受的答案快大约2000倍。

如果您坚持使用正则表达式进行查找,请使用此基于Trie的版本,该版本仍比正则表达式联合快1000倍。

理论

如果您的句子不是笨拙的字符串,每秒处理50个以上的句子可能是可行的。

如果将所有禁止的单词保存到集合中,则可以非常快速地检查该集合中是否包含另一个单词。

将逻辑打包到一个函数中,将此函数作为参数提供给re.sub您,您就完成了!

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)

转换后的句子为:

' .  !
  .
GiraffeElephantBoat
sfgsdg sdwerha aswertwe

注意:

  • 搜索不区分大小写(感谢lower()
  • 用替换一个单词""可能会留下两个空格(如您的代码中所示)
  • 使用python3,\w+还可以匹配带重音符号的字符(例如"ångström")。
  • 任何非单词字符(制表符,空格,换行符,标记等)都将保持不变。

性能

一百万个句子,banned_words近十万个单词,脚本运行时间不到7秒。

相比之下,Liteye的答案需要1万个句子需要160秒。

由于n是单词的总数和m被禁止的单词的数量,OP和Liteye的代码为O(n*m)

相比之下,我的代码应在中运行O(n+m)。考虑到句子比禁止词多得多,该算法变为O(n)

正则表达式联合测试

使用'\b(word1|word2|...|wordN)\b'模式进行正则表达式搜索的复杂性是什么?是O(N)还是O(1)

很难了解正则表达式引擎的工作方式,因此让我们编写一个简单的测试。

此代码将10**i随机的英语单词提取到列表中。它创建相应的正则表达式联合,并用不同的词对其进行测试:

  • 一个人显然不是一个词(以开头#
  • 一个是列表中的第一个单词
  • 一个是列表中的最后一个单词
  • 一个看起来像一个单词,但不是


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))

它输出:

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

因此,看起来像一个带有'\b(word1|word2|...|wordN)\b'模式的单词的搜索具有:

  • O(1) 最好的情况
  • O(n/2) 一般情况,仍然 O(n)
  • O(n) 最糟糕的情况

这些结果与简单的循环搜索一致。

regex联合的一种更快的替代方法是从trie创建regex模式

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.


回答 2

TLDR

如果您想要最快的基于正则表达式的解决方案,请使用此方法。对于类似于OP的数据集,它比接受的答案快大约1000倍。

如果您不关心正则表达式,请使用此基于集合的版本,它比正则表达式联合快2000倍。

使用Trie优化正则表达式

一个简单的正则表达式工会的做法与许多禁用词语变得缓慢,这是因为正则表达式引擎不会做了很好的工作优化格局。

可以使用所有禁止的单词创建Trie并编写相应的正则表达式。生成的trie或regex并不是真正的人类可读的,但是它们确实允许非常快速的查找和匹配。

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

正则表达式联盟

该列表将转换为特里:

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

然后到此正则表达式模式:

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

正则表达式

巨大的优势在于,要测试是否zoo匹配,正则表达式引擎只需比较第一个字符(不匹配),而无需尝试5个单词。这是5个单词的预处理过大杀伤力,但它显示了成千上万个单词的有希望的结果。

请注意,使用(?:)非捕获组是因为:

  • foobar|baz将匹配foobarbaz但不匹配foobaz
  • foo(bar|baz)将不需要的信息保存到捕获组

这是一个经过稍微修改的gist,我们可以将其用作trie.py库:

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())

测试

这是一个小测试(与测试相同):

# 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))

它输出:

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

对于信息,正则表达式开始如下:

(?: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))?)|或(?:(?:\'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)?)|盖尔| 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( ?: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 ...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。 。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。 。

这确实让人难以理解,但是对于100000个禁用词的列表而言,此Trie regex比简单的regex联合快1000倍!

这是完整的trie的图,并通过trie-python-graphviz和graphviz 导出twopi

在此处输入图片说明

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


回答 3

您可能想尝试的一件事是对句子进行预处理以对单词边界进行编码。基本上,通过划分单词边界将每个句子变成单词列表。

这应该更快,因为要处理一个句子,您只需要逐步检查每个单词并检查它是否匹配即可。

当前,正则表达式搜索每次必须再次遍历整个字符串,以查找单词边界,然后在下一次遍历之前“舍弃”这项工作的结果。

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.


回答 4

好吧,这是一个快速简单的解决方案,带有测试仪。

取胜策略:

re.sub(“ \ w +”,repl,sentence)搜索单词。

“ repl”可以是可调用的。我使用了一个执行字典查找的函数,该字典包含要搜索和替换的单词。

这是最简单,最快的解决方案(请参见下面的示例代码中的函数replace4)。

次好的

想法是使用re.split将句子拆分为单词,同时保留分隔符以稍后重建句子。然后,通过简单的字典查找完成替换。

(请参见下面的示例代码中的函数replace3)。

功能示例的时间:

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)

...和代码:

#! /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 )

编辑:检查是否传递小写的句子列表并编辑repl时,您也可以忽略小写

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

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

Winning 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

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.

(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)

回答 5

也许Python不是这里的正确工具。这是Unix工具链中的一个

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

假设您的黑名单文件已经过预处理,并添加了字边界。步骤是:将文件转换为双倍行距,将每个句子拆分为每行一个单词,从文件中批量删除黑名单单词,然后合并回行。

这应该至少快一个数量级。

用于从单词中预处理黑名单文件(每行一个单词)

sed 's/.*/\\b&\\b/' words > blacklist

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

回答 6

这个怎么样:

#!/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)

这些解决方案在单词边界上划分并查找集合中的每个单词。它们应该比re.sub单词替代(Liteyes的解决方案)更快,因为这些解决方案是O(n),其中n是由于amortized O(1)设置查找而导致的,而使用正则表达式替代项将导致regex引擎必须检查单词是否匹配在每个字符上,而不仅仅是在单词边界上。我的解决方案a格外小心,以保留原始文本中使用的空格(即,它不压缩空格,并保留制表符,换行符和其他空格字符),但是如果您决定不关心它,则可以从输出中删除它们应该非常简单。

我在corpus.txt上进行了测试,corpus.txt是从Gutenberg Project下载的多本电子书的串联,并且banned_words.txt是从Ubuntu的单词表(/ usr / share / dict / american-english)中随机选择的20000个单词。处理862462个句子(约占PyPy的一半)大约需要30秒。我已将句子定义为以“。”分隔的任何内容。

$ # 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特别受益于第二种方法,而CPython在第一种方法上表现更好。上面的代码在Python 2和Python 3上都可以使用。

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.


回答 7

实用方法

下述解决方案使用大量内存将所有文本存储在同一字符串中,并降低了复杂度。如果RAM是一个问题,请在使用前三思。

使用join/ split技巧,您可以完全避免循环,从而可以加快算法的速度。

  • 用特殊分隔符连接句子,这些特殊分隔符不包含在句子中:
  • merged_sentences = ' * '.join(sentences)

  • 使用|“或”正则表达式语句为需要从句子中摆脱的所有单词编译一个正则表达式:
  • regex = re.compile(r'\b({})\b'.format('|'.join(words)), re.I) # re.I is a case insensitive flag

  • 用已编译的正则表达式对单词下标,并用特殊的分隔符将其拆分回单独的句子:
  • clean_sentences = re.sub(regex, "", merged_sentences).split(' * ')

    性能

    "".join复杂度为O(n)。这是非常直观的,但是无论如何都会有一个简短的报价来源:

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

    因此,join/split有了O(words)+ 2 * O(sentences)仍然是线性复杂度,而初始方法为2 * O(N 2)。


    顺便说一句,不要使用多线程。GIL将阻止每个操作,因为您的任务严格地受CPU限制,因此GIL没有机会被释放,但是每个线程将同时发送滴答声,这会导致额外的工作量,甚至导致操作达到无穷大。

    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.


    回答 8

    将所有句子连接到一个文档中。使用Aho-Corasick算法的任何实现(这里是)来查找所有“不好”的单词。遍历文件,替换每个坏词,更新后跟的发现词的偏移量等。

    Concatenate all your sentences into one document. Use any implementation of the Aho-Corasick algorithm (here's one) to locate all your "bad" words. Traverse the file, replacing each bad word, updating the offsets of found words that follow etc.


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