下面是一个简短的例子，应该做的技巧与正则表达式:

import re

rep = {"condition1": "", "condition2": "text"} # define desired replacements here

# use these three lines to do the replacement
rep = dict((re.escape(k), v) for k, v in rep.iteritems()) 
#Python 3 renamed dict.iteritems to dict.items so use rep.items() for latest versions
pattern = re.compile("|".join(rep.keys()))
text = pattern.sub(lambda m: rep[re.escape(m.group(0))], text)

例如:

>>> pattern.sub(lambda m: rep[re.escape(m.group(0))], "(condition1) and --condition2--")
'() and --text--'

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)

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'

注意:测试你的案例，见注释。

这里有一个例子，它在长弦上更有效，有许多小的替换。

source = "Here is foo, it does moo!"

replacements = {
    'is': 'was', # replace 'is' with 'was'
    'does': 'did',
    '!': '?'
}

def replace(source, replacements):
    finder = re.compile("|".join(re.escape(k) for k in replacements.keys())) # matches every string we want replaced
    result = []
    pos = 0
    while True:
        match = finder.search(source, pos)
        if match:
            # cut off the part up until match
            result.append(source[pos : match.start()])
            # cut off the matched part and replace it in place
            result.append(replacements[source[match.start() : match.end()]])
            pos = match.end()
        else:
            # the rest after the last match
            result.append(source[pos:])
            break
    return "".join(result)

print replace(source, replacements)

关键是要避免长字符串的多次连接。我们将源字符串切成片段，在我们形成列表时替换一些片段，然后将整个字符串连接回字符串。

你真的不应该这样做，但我觉得这太酷了:

>>> replacements = {'cond1':'text1', 'cond2':'text2'}
>>> cmd = 'answer = s'
>>> for k,v in replacements.iteritems():
>>>     cmd += ".replace(%s, %s)" %(k,v)
>>> exec(cmd)

现在，答案是所有替换的结果

再说一次，这是非常俗气的，不是你应该经常使用的东西。但我很高兴知道如果你需要的话，你可以这样做。