sentence='its some sentence with a something text'

def replaceAll(f,Array1,Array2):
    if len(Array1)==len(Array2):
        for x in range(len(Array1)):
            return f.replace(Array1[x],Array2[x])

newSentence=replaceAll(sentence,['a','sentence','something'],['another','sentence','something something'])

print(newSentence)

这只是F.J和mini夸克的一个更简洁的概述，bgusach的伟大回答和最后但决定性的改进。所有你需要实现多个同步字符串替换是以下函数:

def multiple_replace(string, rep_dict):
    pattern = re.compile("|".join([re.escape(k) for k in sorted(rep_dict,key=len,reverse=True)]), flags=re.DOTALL)
    return pattern.sub(lambda x: rep_dict[x.group(0)], string)

用法:

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

如果您愿意，您可以从这个更简单的函数开始创建自己的专用替换函数。

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。

这是我的0.02美元。它基于Andrew Clark的答案，只是更清楚一点，它还涵盖了当一个字符串被替换为另一个字符串的子字符串时的情况(更长的字符串胜出)

def multireplace(string, replacements):
    """
    Given a string and a replacement map, it returns the replaced string.

    :param str string: string to execute replacements on
    :param dict replacements: replacement dictionary {value to find: value to replace}
    :rtype: str

    """
    # Place longer ones first to keep shorter substrings from matching
    # where the longer ones should take place
    # For instance given the replacements {'ab': 'AB', 'abc': 'ABC'} against 
    # the string 'hey abc', it should produce 'hey ABC' and not 'hey ABc'
    substrs = sorted(replacements, key=len, reverse=True)

    # Create a big OR regex that matches any of the substrings to replace
    regexp = re.compile('|'.join(map(re.escape, substrs)))

    # For each match, look up the new string in the replacements
    return regexp.sub(lambda match: replacements[match.group(0)], string)

这就是这个要点，如果你有任何建议，请随意修改。

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!: -)

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.'