[英]Regex to replace %variables%
I've been yanking clumps of hair out for 30 minutes doing this one...我一直在拉头发30分钟做这个......
I have a dictionary, like so:我有一本字典,像这样:
{'search': 'replace',
'foo': 'bar'}
And a string like this:还有这样的字符串:
Foo bar %foo% % search %.
I'd like to replace each variable with it's equivalent text from the dictionary:我想用字典中的等效文本替换每个变量:
Foo bar bar replace.
My current regex fails, so here it is ( key
and value
are from dictionary.items()
):我当前的正则表达式失败,所以它是( key
和value
来自dictionary.items()
):
re.sub(r'%\d+' + key + '[^%]\d+%', value, text)
Any help would be appreciated, as this regex stuff is driving me nuts...任何帮助将不胜感激,因为这个正则表达式的东西让我发疯......
If you're flexible with your syntax in your string, Python has a built in mechanism for that:如果您对字符串中的语法很灵活,Python 有一个内置机制:
>>> print 'Hello, %(your_name)s, my name is %(my_name)s' % {'your_name': 'Blender', 'my_name': 'Ken'}
Hello, Blender, my name is Ken
Alternatively, if you want that syntax, I'd avoid regular expressions and just do this:或者,如果你想要那种语法,我会避免使用正则表达式,而是这样做:
>>> vars = {'search': 'replace',
... 'foo': 'bar'}
>>> mystring = "Foo bar %foo% % search %."
>>> for k, v in vars.items():
... mystring = mystring.replace('%%%s%%' % k, v)
...
>>> print mystring
Foo bar bar % search %.
If you want it in one statement, you could do the following (assuming s is the string and d is the dictionary):如果您想在一个语句中使用它,您可以执行以下操作(假设 s 是字符串,d 是字典):
re.sub(r"[%]\s*[^%]+\s*[%]",lambda k:d.get(k[1,-1].strip(),k),s)
This uses a function in the replacement part to get each value from the dictionary, and ignores if it is not in the dictionary.这在替换部分使用 function 从字典中获取每个值,如果不在字典中则忽略。
Edit: fixed bug with unwanted whitespace appearing in lookup key编辑:修复了查找键中出现不需要的空格的错误
Maybe I'm missing something, but wouldn't the following regex just work?也许我遗漏了一些东西,但下面的正则表达式不会起作用吗?
re.sub(r'%\s?' + key + '\s?%', value, text)
The only thing that's a bit special are the optional spaces;唯一有点特别的是可选空格; they can be matched with \s?
它们可以与\s?
. .
Using the replacement function support of re.sub
:使用re.sub
的替换 function 支持:
def replace(s, kw, pattern=re.compile(r'%\s*(\w+)\s*%')):
"""
Replace delimited keys in a string with values from kw.
Any pattern may be used, as long as the first group defines the lookup key.
"""
lookup = lambda match: kw.get(match.group(1), match.group())
return pattern.sub(lookup, s)
>>> replace('Foo bar %foo% % search %.', {'search': 'replace', 'foo': 'bar'})
'Foo bar bar replace.'
You can change the lookup
function to customize how lookup errors are treated, and so on.您可以更改lookup
function 以自定义如何处理查找错误,等等。
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