python 与python中php的str_replace类似的功能？
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similar function to php's str_replace in python?
提问by Mohamed
is there a similar function in python that takes search(array) and replace(array) as a parameter? Then takes a value from each array and uses them to do search and replace on subject(string).
python中是否有类似的函数将search(array)和replace(array)作为参数？然后从每个数组中获取一个值并使用它们对主题（字符串）进行搜索和替换。
I know I can achieve this using for loops, but just looking more elegant way.
我知道我可以使用 for 循环来实现这一点，但只是看起来更优雅。
回答by John Fouhy
I believe the answer is no.
我相信答案是否定的。
I would specify your search/replace strings in a list, and the iterate over it:
我会在列表中指定您的搜索/替换字符串，并对其进行迭代：
edits = [(search0, replace0), (search1, replace1), (search2, replace2)] # etc.
for search, replace in edits:
s = s.replace(search, replace)
Even if python did have a str_replace-style function, I think I would still separate out my search/replace strings as a list, so really this is only taking one extra line of code.
即使 python 确实有一个str_replace-style 函数，我想我仍然会将我的搜索/替换字符串作为一个列表分开，所以这实际上只需要多一行代码。
Finally, this is a programming language after all. If it doesn't supply the function you want, you can always define it yourself.
最后，这毕竟是一种编程语言。如果它没有提供你想要的功能，你总是可以自己定义它。
回答by Anon
Heh - you could use the one-liner below whose elegance is second only to its convenience :-P
呵呵 - 你可以使用下面的单线，它的优雅仅次于它的方便:-P
(Acts like PHP when search is longer than replace, too, if I read that correctly in the PHP docs.):
（如果我在 PHP 文档中正确阅读了这一点，那么当搜索时间长于替换时，行为也像 PHP。）：
**** Edit: This new version works for all sized substrings to replace. ****
**** 编辑：这个新版本适用于所有大小的子字符串来替换。****
>>> subject = "Coming up with these convoluted things can be very addictive."
>>> search = ['Coming', 'with', 'things', 'addictive.', ' up', ' these', 'convoluted ', ' very']
>>> replace = ['Making', 'Python', 'one-liners', 'fun!']
>>> reduce(lambda s, p: s.replace(p[0],p[1]),[subject]+zip(search, replace+['']*(len(search)-len(replace))))
'Making Python one-liners can be fun!'
回答by Glenn Maynard
Do it with regexps:
用正则表达式来做：
import re
def replace_from_list(replacements, str):
def escape_string_to_regex(str):
return re.sub(r"([\.^$*+?{}[\]|\(\)])", r"\", str)
def get_replacement(match):
return replacements[match.group(0)]
replacements = dict(replacements)
replace_from = [escape_string_to_regex(r) for r in replacements.keys()]
regex = "|".join(["(%s)" % r for r in replace_from])
repl = re.compile(regex)
return repl.sub(get_replacement, str)
# Simple replacement:
assert replace_from_list([("in1", "out1")], "in1") == "out1"
# Replacements are never themselves replaced, even if later search strings match
# earlier destination strings:
assert replace_from_list([("1", "2"), ("2", "3")], "123") == "233"
# These are plain strings, not regexps:
assert replace_from_list([("...", "out")], "abc ...") == "abc out"
Using regexps for this makes the searching fast. This won't iteratively replace replacements with further replacements, which is usuallywhat's wanted.
为此使用正则表达式可以使搜索快速。这不会用进一步的替换迭代替换替换，这通常是我们想要的。
回答by Юки Эндо
Made a tiny recursive function for this
为此做了一个很小的递归函数
def str_replace(sbjct, srch, rplc):
if len(sbjct) == 0:
return ''
if len(srch) == 1:
return sbjct.replace(srch[0], rplc[0])
lst = sbjct.split(srch[0])
reslst = []
for s in lst:
reslst.append(str_replace(s, srch[1:], rplc[1:]))
return rplc[0].join(reslst);
