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)

这里有一个使用reduce的第一个解决方案的变体，如果你喜欢功能性的。：）

repls = {'hello' : 'goodbye', 'world' : 'earth'}
s = 'hello, world'
reduce(lambda a, kv: a.replace(*kv), repls.iteritems(), s)

马蒂诺的版本更好:

repls = ('hello', 'goodbye'), ('world', 'earth')
s = 'hello, world'
reduce(lambda a, kv: a.replace(*kv), repls, s)

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.'

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.

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。

我不知道速度如何，但这是我日常的快速解决方法:

reduce(lambda a, b: a.replace(*b)
    , [('o','W'), ('t','X')] #iterable of pairs: (oldval, newval)
    , 'tomato' #The string from which to replace values
    )

．.． 但我喜欢上面的#1正则表达式答案。注意:如果一个新值是另一个值的子字符串，那么该操作是不可交换的。