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

我把这句话建立在fj的精彩回答上:

import re

def multiple_replacer(*key_values):
    replace_dict = dict(key_values)
    replacement_function = lambda match: replace_dict[match.group(0)]
    pattern = re.compile("|".join([re.escape(k) for k, v in key_values]), re.M)
    return lambda string: pattern.sub(replacement_function, string)

def multiple_replace(string, *key_values):
    return multiple_replacer(*key_values)(string)

一针用法:

>>> replacements = (u"café", u"tea"), (u"tea", u"café"), (u"like", u"love")
>>> print multiple_replace(u"Do you like café? No, I prefer tea.", *replacements)
Do you love tea? No, I prefer café.

注意，由于替换只在一次传递中完成，“café”会变成“tea”，但不会变回“café”。

如果你需要做相同的替换多次，你可以很容易地创建一个替换函数:

>>> my_escaper = multiple_replacer(('"','\\"'), ('\t', '\\t'))
>>> many_many_strings = (u'This text will be escaped by "my_escaper"',
                       u'Does this work?\tYes it does',
                       u'And can we span\nmultiple lines?\t"Yes\twe\tcan!"')
>>> for line in many_many_strings:
...     print my_escaper(line)
... 
This text will be escaped by \"my_escaper\"
Does this work?\tYes it does
And can we span
multiple lines?\t\"Yes\twe\tcan!\"

改进:

将代码转换为函数 增加了多线支持 修正了逃跑的错误 容易创建一个函数，用于特定的多个替换

享受吧!: -)

你可以做一个漂亮的循环函数。

def replace_all(text, dic):
    for i, j in dic.iteritems():
        text = text.replace(i, j)
    return text

其中text是完整的字符串，dic是字典-每个定义都是一个字符串，将替换与术语匹配的字符串。

注意:在Python 3中，iteritems()已被items()取代

注意:Python字典没有迭代的可靠顺序。此解决方案仅在以下情况下解决您的问题:

替换的顺序无关紧要 替换者可以改变之前替换者的结果

更新:上述与插入顺序相关的语句不适用于大于或等于3.6的Python版本，因为标准字典已更改为使用插入顺序进行迭代。

例如:

d = { "cat": "dog", "dog": "pig"}
my_sentence = "This is my cat and this is my dog."
replace_all(my_sentence, d)
print(my_sentence)

可能输出#1:

"This is my pig and this is my pig."

可能的输出#2

"This is my dog and this is my pig."

一个可能的解决方法是使用OrderedDict。

from collections import OrderedDict
def replace_all(text, dic):
    for i, j in dic.items():
        text = text.replace(i, j)
    return text
od = OrderedDict([("cat", "dog"), ("dog", "pig")])
my_sentence = "This is my cat and this is my dog."
replace_all(my_sentence, od)
print(my_sentence)

输出:

"This is my pig and this is my pig."

注意事项#2:如果你的文本字符串太大或字典中有很多对，效率就会很低。

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

在我的情况下，我需要一个简单的唯一键替换名称，所以我想到了这个:

a = 'This is a test string.'
b = {'i': 'I', 's': 'S'}
for x,y in b.items():
    a = a.replace(x, y)
>>> a
'ThIS IS a teSt StrIng.'

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'