下面是一个支持基本正则表达式替换的版本。主要的限制是表达式不能包含子组，并且可能存在一些边缘情况:

基于@bgusach和其他的代码

import re

class StringReplacer:

    def __init__(self, replacements, ignore_case=False):
        patterns = sorted(replacements, key=len, reverse=True)
        self.replacements = [replacements[k] for k in patterns]
        re_mode = re.IGNORECASE if ignore_case else 0
        self.pattern = re.compile('|'.join(("({})".format(p) for p in patterns)), re_mode)
        def tr(matcher):
            index = next((index for index,value in enumerate(matcher.groups()) if value), None)
            return self.replacements[index]
        self.tr = tr

    def __call__(self, string):
        return self.pattern.sub(self.tr, string)

测试

table = {
    "aaa"    : "[This is three a]",
    "b+"     : "[This is one or more b]",
    r"<\w+>" : "[This is a tag]"
}

replacer = StringReplacer(table, True)

sample1 = "whatever bb, aaa, <star> BBB <end>"

print(replacer(sample1))

# output: 
# whatever [This is one or more b], [This is three a], [This is a tag] [This is one or more b] [This is a tag]

诀窍是通过位置来识别匹配的组。它不是超级高效(O(n))，但它是有效的。

index = next((index for index,value in enumerate(matcher.groups()) if value), None)

替换是一次完成的。

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']])

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。

对于只替换一个字符，使用翻译和str.maketrans是我最喜欢的方法。

Tl;dr > result_string = your_string.translate(str.maketrans(dict_mapping))

demo

my_string = 'This is a test string.'
dict_mapping = {'i': 's', 's': 'S'}
result_good = my_string.translate(str.maketrans(dict_mapping))
result_bad = my_string
for x, y in dict_mapping.items():
    result_bad = result_bad.replace(x, y)
print(result_good)  # ThsS sS a teSt Strsng.
print(result_bad)   # ThSS SS a teSt StrSng.

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--'