这是我对这个问题的解决办法。我把它用在聊天机器人上，一次替换不同的单词。

def mass_replace(text, dct):
    new_string = ""
    old_string = text
    while len(old_string) > 0:
        s = ""
        sk = ""
        for k in dct.keys():
            if old_string.startswith(k):
                s = dct[k]
                sk = k
        if s:
            new_string+=s
            old_string = old_string[len(sk):]
        else:
            new_string+=old_string[0]
            old_string = old_string[1:]
    return new_string

print mass_replace("The dog hunts the cat", {"dog":"cat", "cat":"dog"})

这就成了猫捉狗

从安德鲁的宝贵答案开始，我开发了一个脚本，从一个文件加载字典，并详细说明所有文件上打开的文件夹做替换。脚本从一个外部文件加载映射，您可以在该文件中设置分隔符。我是一个初学者，但我发现这个脚本在多个文件中做多个替换时非常有用。它在几秒钟内加载了一个包含1000多个条目的字典。这并不优雅，但对我来说很管用

import glob
import re

mapfile = input("Enter map file name with extension eg. codifica.txt: ")
sep = input("Enter map file column separator eg. |: ")
mask = input("Enter search mask with extension eg. 2010*txt for all files to be processed: ")
suff = input("Enter suffix with extension eg. _NEW.txt for newly generated files: ")

rep = {} # creation of empy dictionary

with open(mapfile) as temprep: # loading of definitions in the dictionary using input file, separator is prompted
    for line in temprep:
        (key, val) = line.strip('\n').split(sep)
        rep[key] = val

for filename in glob.iglob(mask): # recursion on all the files with the mask prompted

    with open (filename, "r") as textfile: # load each file in the variable text
        text = textfile.read()

        # start replacement
        #rep = dict((re.escape(k), v) for k, v in rep.items()) commented to enable the use in the mapping of re reserved characters
        pattern = re.compile("|".join(rep.keys()))
        text = pattern.sub(lambda m: rep[m.group(0)], text)

        #write of te output files with the prompted suffice
        target = open(filename[:-4]+"_NEW.txt", "w")
        target.write(text)
        target.close()

您可以使用pandas库和replace函数，它既支持精确匹配，也支持正则表达式替换。例如:

df = pd.DataFrame({'text': ['Billy is going to visit Rome in November', 'I was born in 10/10/2010', 'I will be there at 20:00']})

to_replace=['Billy','Rome','January|February|March|April|May|June|July|August|September|October|November|December', '\d{2}:\d{2}', '\d{2}/\d{2}/\d{4}']
replace_with=['name','city','month','time', 'date']

print(df.text.replace(to_replace, replace_with, regex=True))

修改后的文本为:

0    name is going to visit city in month
1                      I was born in date
2                 I will be there at time

你可以在这里找到一个例子。请注意，文本上的替换是按照它们在列表中出现的顺序进行的

我的方法是首先将字符串标记化，然后决定每个标记是否包含它。

潜在地，如果我们可以假设一个hashmap/set的O(1)查找，可能会更好:

remove_words = {"we", "this"}
target_sent = "we should modify this string"
target_sent_words = target_sent.split()
filtered_sent = " ".join(list(filter(lambda word: word not in remove_words, target_sent_words)))

Filtered_sent现在是'应该修改字符串'

我不知道速度如何，但这是我日常的快速解决方法:

reduce(lambda a, b: a.replace(*b)
    , [('o','W'), ('t','X')] #iterable of pairs: (oldval, newval)
    , 'tomato' #The string from which to replace values
    )

．.． 但我喜欢上面的#1正则表达式答案。注意:如果一个新值是另一个值的子字符串，那么该操作是不可交换的。

这里有一个使用reduce的第一个解决方案的变体，如果你喜欢功能性的。：）

repls = {'hello' : 'goodbye', 'world' : 'earth'}
s = 'hello, world'
reduce(lambda a, kv: a.replace(*kv), repls.iteritems(), s)

马蒂诺的版本更好:

repls = ('hello', 'goodbye'), ('world', 'earth')
s = 'hello, world'
reduce(lambda a, kv: a.replace(*kv), repls, s)