我需要一个解决方案，其中字符串可以被替换为正则表达式， 例如，通过将多个空格字符替换为一个空格字符来帮助规范化长文本。根据其他人(包括MiniQuark和mmj)的一系列答案，我得出了以下结论:

def multiple_replace(string, reps, re_flags = 0):
    """ Transforms string, replacing keys from re_str_dict with values.
    reps: dictionary, or list of key-value pairs (to enforce ordering;
          earlier items have higher priority).
          Keys are used as regular expressions.
    re_flags: interpretation of regular expressions, such as re.DOTALL
    """
    if isinstance(reps, dict):
        reps = reps.items()
    pattern = re.compile("|".join("(?P<_%d>%s)" % (i, re_str[0])
                                  for i, re_str in enumerate(reps)),
                         re_flags)
    return pattern.sub(lambda x: reps[int(x.lastgroup[1:])][1], string)

它适用于其他答案中给出的例子，例如:

>>> multiple_replace("(condition1) and --condition2--",
...                  {"condition1": "", "condition2": "text"})
'() and --text--'

>>> multiple_replace('hello, world', {'hello' : 'goodbye', 'world' : 'earth'})
'goodbye, earth'

>>> multiple_replace("Do you like cafe? No, I prefer tea.",
...                  {'cafe': 'tea', 'tea': 'cafe', 'like': 'prefer'})
'Do you prefer tea? No, I prefer cafe.'

对我来说，最重要的是你也可以使用正则表达式，例如只替换整个单词，或规范化空白:

>>> s = "I don't want to change this name:\n  Philip II of Spain"
>>> re_str_dict = {r'\bI\b': 'You', r'[\n\t ]+': ' '}
>>> multiple_replace(s, re_str_dict)
"You don't want to change this name: Philip II of Spain"

如果你想使用字典键作为普通字符串， 你可以在调用multiple_replace之前转义这些，例如使用下面的函数:

def escape_keys(d):
    """ transform dictionary d by applying re.escape to the keys """
    return dict((re.escape(k), v) for k, v in d.items())

>>> multiple_replace(s, escape_keys(re_str_dict))
"I don't want to change this name:\n  Philip II of Spain"

下面的函数可以帮助在你的字典键中找到错误的正则表达式(因为来自multiple_replace的错误消息不是很明显):

def check_re_list(re_list):
    """ Checks if each regular expression in list is well-formed. """
    for i, e in enumerate(re_list):
        try:
            re.compile(e)
        except (TypeError, re.error):
            print("Invalid regular expression string "
                  "at position {}: '{}'".format(i, e))

>>> check_re_list(re_str_dict.keys())

请注意，它没有链接替换，而是同时执行它们。这使得它更有效率，而不会限制它能做什么。为了模仿链接的效果，你可能只需要添加更多的字符串替换对，并确保这些对的预期顺序:

>>> multiple_replace("button", {"but": "mut", "mutton": "lamb"})
'mutton'
>>> multiple_replace("button", [("button", "lamb"),
...                             ("but", "mut"), ("mutton", "lamb")])
'lamb'

下面是另一种使用字典的方法:

listA="The cat jumped over the house".split()
modify = {word:word for number,word in enumerate(listA)}
modify["cat"],modify["jumped"]="dog","walked"
print " ".join(modify[x] for x in listA)

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'

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

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'