我想建议使用字符串模板。只需将要替换的字符串放在字典中，一切就都设置好了!示例来自docs.python.org

>>> from string import Template
>>> s = Template('$who likes $what')
>>> s.substitute(who='tim', what='kung pao')
'tim likes kung pao'
>>> d = dict(who='tim')
>>> Template('Give $who $100').substitute(d)
Traceback (most recent call last):
[...]
ValueError: Invalid placeholder in string: line 1, col 10
>>> Template('$who likes $what').substitute(d)
Traceback (most recent call last):
[...]
KeyError: 'what'
>>> Template('$who likes $what').safe_substitute(d)
'tim likes $what'

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()

为什么没有这样的解决方案呢?

s = "The quick brown fox jumps over the lazy dog"
for r in (("brown", "red"), ("lazy", "quick")):
    s = s.replace(*r)

#output will be:  The quick red fox jumps over the quick dog

我今天也面临类似的问题，我不得不多次使用.replace()方法，但这对我来说感觉不太好。所以我做了这样的事情:

REPLACEMENTS = {'<': '&lt;', '>': '&gt;', '&': '&amp;'}

event_title = ''.join([REPLACEMENTS.get(c,c) for c in event['summary']])

从Python 3.8开始，并引入赋值表达式(PEP 572)(:=运算符)，我们可以在一个列表理解式中应用替换:

# text = "The quick brown fox jumps over the lazy dog"
# replacements = [("brown", "red"), ("lazy", "quick")]
[text := text.replace(a, b) for a, b in replacements]
# text = 'The quick red fox jumps over the quick dog'

我在学校作业中也做过类似的练习。这就是我的解

dictionary = {1: ['hate', 'love'],
              2: ['salad', 'burger'],
              3: ['vegetables', 'pizza']}

def normalize(text):
    for i in dictionary:
        text = text.replace(dictionary[i][0], dictionary[i][1])
    return text

自己查看测试字符串上的结果

string_to_change = 'I hate salad and vegetables'
print(normalize(string_to_change))