Question

给定一个字符串和一个应该替换为占位符的子字符串列表，例如

import re
from copy import copy 

phrases = ["'s morgen", "'s-Hertogenbosch", "depository financial institution"]
original_text = "Something, 's morgen, ik 's-Hertogenbosch im das depository financial institution gehen"

第一个目标是首先使用索引占位符替换phrases中original_text的子字符串，例如：

text = copy(original_text)
backplacement = {}
for i, phrase in enumerate(phrases):
    backplacement["MWEPHRASE{}".format(i)] = phrase.replace(' ', '_')
    text = re.sub(r"{}".format(phrase), "MWEPHRASE{}".format(i), text)
print(text)

[OUT]：

Something, MWEPHRASE0, ik MWEPHRASE1 im das MWEPHRASE2 gehen

然后，有一些函数可以使用占位符操作text，例如

cleaned_text = func('Something, MWEPHRASE0, ik MWEPHRASE1 im das MWEPHRASE2 gehen')
print(cleaned_text)

输出：

MWEPHRASE0 ik MWEPHRASE1 MWEPHRASE2

最后一步是以倒退的方式进行替换，并放回原来的短语，即

' '.join([backplacement[tok] if tok in backplacement else tok for tok in clean_text.split()])

[OUT]：

"'s_morgen ik 's-Hertogenbosch depository_financial_institution"

问题是：

如果phrases中的子项列表很大，那么进行第一次替换和最后一次替换的时间将花费很长时间。

有没有办法用正则表达式进行替换/替换？

使用re.sub(r"{}".format(phrase), "MWEPHRASE{}".format(i), text)正则表达式替换并非特别有用。如果短语中的子串不匹配完整的单词，

E.g。

phrases = ["org", "'s-Hertogenbosch", "depository financial institution"]
original_text = "Something, 's morgen, ik 's-Hertogenbosch im das depository financial institution gehen"
backplacement = {}
text = copy(original_text)
for i, phrase in enumerate(phrases):
    backplacement["MWEPHRASE{}".format(i)] = phrase.replace(' ', '_')
    text = re.sub(r"{}".format(phrase), "MWEPHRASE{}".format(i), text)
print(text)

我们得到一个尴尬的输出：

Something, 's mMWEPHRASE0en, ik MWEPHRASE1 im das MWEPHRASE2 gehen

我已尝试使用'\b{}\b'.format(phrase)，但对于带标点符号的短语不起作用，即

phrases = ["'s morgen", "'s-Hertogenbosch", "depository financial institution"]
original_text = "Something, 's morgen, ik 's-Hertogenbosch im das depository financial institution gehen"
backplacement = {}
text = copy(original_text)
for i, phrase in enumerate(phrases):
    backplacement["MWEPHRASE{}".format(i)] = phrase.replace(' ', '_')
    text = re.sub(r"\b{}\b".format(phrase), "MWEPHRASE{}".format(i), text)
print(text)

[OUT]：

Something, 's morgen, ik 's-Hertogenbosch im das MWEPHRASE2 gehen

是否有一些地方可以表示re.sub正则表达式模式中短语的单词边界？

Answer 1

您可以拆分它而不是使用re.sub。

def do_something_with_str(string):
    # do something with string here.
    # for example let's wrap the string with "@" symbol if it's not empty
    return f"@{string}" if string else string


def get_replaced_list(string, words):
    result = [(string, True), ]

    # we take each word we want to replace
    for w in words:

        new_result = []

        # Getting each word in old result
        for r in result:

            # Now we split every string in results using our word.
            split_list = list((x, True) for x in r[0].split(w)) if r[1] else list([r, ])

            # If we replace successfully - add all the strings
            if len(split_list) > 1:

                # This one would be for [text, replaced, text, replaced...]
                sub_result = []
                ws = [(w, False), ] * (len(split_list) - 1)
                for x, replaced in zip(split_list, ws):
                    sub_result.append(x)
                    sub_result.append(replaced)
                sub_result.append(split_list[-1])

                # Add to new result
                new_result.extend(sub_result)

            # If not - just add it to results
            else:
                new_result.extend(split_list)
        result = new_result
    return result


if __name__ == '__main__':
    initial_string = 'acbbcbbcacbbcbbcacbbcbbca'
    words_to_replace = ('a', 'c')
    replaced_list = get_replaced_list(initial_string, words_to_replace)
    modified_list = [(do_something_with_str(x[0]), True) if x[1] else x for x in replaced_list]
    final_string = ''.join([x[0] for x in modified_list])

以上示例的以下变量值：

initial_string = 'acbbcbbcacbbcbbcacbbcbbca'
words_to_replace = ('a', 'c')
replaced_list = [('', True), ('a', False), ('', True), ('c', False), ('bb', True), ('c', False), ('bb', True), ('c', False), ('', True), ('a', False), ('', True), ('c', False), ('bb', True), ('c', False), ('bb', True), ('c', False), ('', True), ('a', False), ('', True), ('c', False), ('bb', True), ('c', False), ('bb', True), ('c', False), ('', True), ('a', False), ('', True)]
modified_list = [('', True), ('a', False), ('', True), ('c', False), ('@bb', True), ('c', False), ('@bb', True), ('c', False), ('', True), ('a', False), ('', True), ('c', False), ('@bb', True), ('c', False), ('@bb', True), ('c', False), ('', True), ('a', False), ('', True), ('c', False), ('@bb', True), ('c', False), ('@bb', True), ('c', False), ('', True), ('a', False), ('', True)]
final_string = 'ac@bbc@bbcac@bbc@bbcac@bbc@bbca'

如您所见，列表包含元组。它们包含两个值 - some string和boolean，表示它是文本还是替换值（文本时为True）。获得替换列表后，您可以像在示例中一样修改它，检查它是否为文本值（if x[1] == True）。希望有所帮助！

P.S。像 f"some string here {some_variable_here}" 这样的字符串格式需要Python 3.6

Answer 2

我认为在此任务中使用正则表达式有两个关键：

使用自定义边界，捕获它们，然后将它们替换为短语。
使用函数在两个方向上处理替换匹配。

以下是使用此方法的实现。我稍微调整了你的文字，重复其中一个短语。

import re
from copy import copy 

original_text = "Something, 's morgen, ik 's-Hertogenbosch im das depository financial institution gehen 's morgen"
text = copy(original_text)

#
# The phrases of interest
#
phrases = ["'s morgen", "'s-Hertogenbosch", "depository financial institution"]

#
# Create the mapping dictionaries
#
phrase_to_mwe = {}
mwe_to_phrase = {}

#
# Build the mappings
#
for i, phrase in enumerate(phrases):

    mwephrase                = "MWEPHRASE{}".format(i)
    mwe_to_phrase[mwephrase] = phrase.replace(' ', '_')
    phrase_to_mwe[phrase]    = mwephrase

#
# Regex match handlers
#
def handle_forward(match):

    b1     = match.group(1)
    phrase = match.group(2)
    b2     = match.group(3)

    return b1 + phrase_to_mwe[phrase] + b2


def handle_backward(match):

    return mwe_to_phrase[match.group(1)]

#
# The forward regex will look like:
#
#    (^|[ ])('s morgen|'s-Hertogenbosch|depository financial institution)([, ]|$)
# 
# which captures three components:
#
#    (1) Front boundary
#    (2) Phrase
#    (3) Back boundary
#
# Anchors allow matching at the beginning and end of the text. Addtional boundary characters can be
# added as necessary, e.g. to allow semicolons after a phrase, we could update the back boundary to:
#
#    ([,; ]|$)
#
regex_forward  = re.compile(r'(^|[ ])(' + '|'.join(phrases) + r')([, ]|$)')
regex_backward = re.compile(r'(MWEPHRASE\d+)')

#
# Pretend we cleaned the text in the middle
#
cleaned = 'MWEPHRASE0 ik MWEPHRASE1 MWEPHRASE2 MWEPHRASE0'

#
# Do the translations
#
text1 = regex_forward .sub(handle_forward,  text)
text2 = regex_backward.sub(handle_backward, cleaned)

print('original: {}'.format(original_text))
print('text1   : {}'.format(text1))
print('text2   : {}'.format(text2))

运行此会生成：

original: Something, 's morgen, ik 's-Hertogenbosch im das depository financial institution gehen 's morgen
text1   : Something, MWEPHRASE0, ik MWEPHRASE1 im das MWEPHRASE2 gehen MWEPHRASE0
text2   : 's_morgen ik 's-Hertogenbosch depository_financial_institution 's_morgen

Answer 3

这是您可以使用的策略：

phrases = ["'s morgen", "'s-Hertogenbosch", "depository financial institution"]
original_text = "Something, 's morgen, ik 's-Hertogenbosch im das depository financial institution gehen"

# need this module for the reduce function
import functools as fn

#convert phrases into a dictionary of numbered placeholders (tokens)
tokens = { kw:"MWEPHRASE%s"%i for i,kw in enumerate(phrases) }

#replace embedded phrases with their respective token
tokenized = fn.reduce(lambda s,kw: tokens[kw].join(s.split(kw)), phrases, original_text)

#Apply text cleaning logic on the tokenized text 
#This assumes the placeholders are left untouched, 
#although it's ok to move them around)
cleaned_text = cleanUpfunction(tokenized)

#reverse the token dictionary (to map original phrases to numbered placeholders)
unTokens = {v:k for k,v in tokens.items() }

#rebuild phrases with original text associated to each token (placeholder)
final_text = fn.reduce(lambda s,kw: unTokens[kw].join(s.split(kw)), phrases, cleaned_text)

Answer 4

您正在寻找的是“多字符串搜索”或“多模式搜索”。更常见的解决方案是Aho-Corasick和Rabin-Karp算法。如果你想自己强迫它，那就选择Rabin-Karp，因为它更容易掌握。否则，你会找到一些库。这是库https://pypi.python.org/pypi/py_aho_corasick的解决方案。

让

phrases = ["'s morgen", "'s-Hertogenbosch", "depository financial institution"]
original_text = "Something, 's morgen, ik 's-Hertogenbosch im das depository financial institution gehen"

并且，出于测试目的：

def clean(text):
    """A simple stub"""
    assert text == 'Something, MWEPHRASE0, ik MWEPHRASE1 im das MWEPHRASE2 gehen'
    return "MWEPHRASE0 ik MWEPHRASE1 MWEPHRASE2"

现在，你必须定义两个自动机，一个用于向外旅程，另一个用于返回。自动机由（键，值）列表定义：

fore_automaton = py_aho_corasick.Automaton([(phrase,"MWEPHRASE{}".format(i)) for i, phrase in enumerate(phrases)])
back_automaton = py_aho_corasick.Automaton([("MWEPHRASE{}".format(i), phrase.replace(' ','_')) for i, phrase in enumerate(phrases)])

自动机将扫描文本并返回匹配列表。匹配是三元组（位置，键，值）。通过对匹配进行一些工作，您将能够通过值替换键：

def process(automaton, text):
    """Returns a new text, with keys of the automaton replaced by values"""
    matches = automaton.get_keywords_found(text.lower()) # text.lower() because auomaton of py_aho_corasick uses lowercase for keys
    bk_value_eks = [(i,v,i+len(k)) for i,k,v in matches] # (begin of key, value, end of key)
    chunks = [bk_value_ek1[1]+text[bk_value_ek1[2]:bk_value_ek2[0]] for bk_value_ek1,bk_value_ek2 in zip([(-1,"",0)]+bk_value_eks, bk_value_eks+[(len(text),"",-1)] if bk_value_ek1[2] <= bk_value_ek2[0]] # see below
    return "".join(chunks)

关于chunks = [bk_value_ek1[1]+text[bk_value_ek1[2]:bk_value_ek2[0]] for bk_value_ek1,bk_value_ek2 in zip([(-1,"",0)]+bk_value_eks, bk_value_eks+[(len(text),"",-1)] if bk_value_ek1[2] <= bk_value_ek2[0]]的简要说明。我几乎像往常一样压缩自己的匹配：zip(arr, arr[1:])将输出(arr[0], arr[1)), (arr[1], arr[2]), ...来考虑与其成功的每场比赛。我在这里放了两个哨兵处理比赛的开始和结束。

对于正常情况，我只输出值（= bk_value_ek1[1]）以及键的结尾和下一个键的开头（text[bk_value_ek1[2]:bk_value_ek2[0]）之间的文本。
begin sentinel有一个空值，其键结束于位置0，因此第一个块将是“”+ text [0：key1的开头]，即第一个键之前的文本。
同样，end sentinel也有一个空值，它的键从文本末尾开始，因此最后一个chunk将是：最后一个匹配的值+ text [最后一个键的结尾：len（text）]

密钥重叠时会发生什么？举个例子：text="abcdef"，phrases={"bcd":"1", "cde":"2"}。您有两个匹配项：(1, "bcd", "1")和(2, "cde", "3")。我们走吧：bk_value_eks = [(1, "1", 4), (2, "2", 5)]。因此，如果没有if bk_value_ek1[2] <= bk_value_ek2[0]，则文本将替换为text[:1]+"1"+text[4:2]+"2"+text[5:]，这是"a"+"1"+""+"2"+"f" = "a12f"而不是"a1ef"（忽略第二场比赛）......

现在，看看结果：

print(process(back_automaton, clean(process(fore_automaton, original_text))))
# "'s_morgen ik 's-Hertogenbosch depository_financial_institution"

您不必为返回定义新的process函数，只需将其赋予back_automaton即可完成此任务。

用占位符替换字符串并在函数后替换它们。

4 个答案: