Question

任何时候我想替换一段文字，这段文字都是大文字的一部分，我总是要这样做：

"(?P<start>some_pattern)(?P<replace>foo)(?P<end>end)"

然后将start组与replace的新数据以及end组的新数据连接起来。

有更好的方法吗？

Answer 1

>>> import re
>>> s = "start foo end"
>>> s = re.sub("foo", "replaced", s)
>>> s
'start replaced end'
>>> s = re.sub("(?<= )(.+)(?= )", lambda m: "can use a callable for the %s text too" % m.group(1), s)
>>> s
'start can use a callable for the replaced text too end'
>>> help(re.sub)
Help on function sub in module re:

sub(pattern, repl, string, count=0)
    Return the string obtained by replacing the leftmost
    non-overlapping occurrences of the pattern in string by the
    replacement repl.  repl can be either a string or a callable;
    if a callable, it's passed the match object and must return
    a replacement string to be used.

Answer 2

在Python re documentation中查找前瞻(?=...)和lookbehinds (?<=...) - 我很确定它们就是你想要的。它们匹配字符串，但不“消耗”它们匹配的字符串的位。

Answer 3

简短版本是在使用Python的re模块的外观中无法使用可变宽度模式。没有办法改变这个：

>>> import re
>>> re.sub("(?<=foo)bar(?=baz)", "quux", "foobarbaz")
'fooquuxbaz'
>>> re.sub("(?<=fo+)bar(?=baz)", "quux", "foobarbaz")

Traceback (most recent call last):
  File "<pyshell#2>", line 1, in <module>
    re.sub("(?<=fo+)bar(?=baz)", "quux", string)
  File "C:\Development\Python25\lib\re.py", line 150, in sub
    return _compile(pattern, 0).sub(repl, string, count)
  File "C:\Development\Python25\lib\re.py", line 241, in _compile
    raise error, v # invalid expression
error: look-behind requires fixed-width pattern

这意味着你需要解决它，最简单的解决方案与你现在正在做的非常相似：

>>> re.sub("(fo+)bar(?=baz)", "\\1quux", "foobarbaz")
'fooquuxbaz'
>>>
>>> # If you need to turn this into a callable function:
>>> def replace(start, replace, end, replacement, search):
        return re.sub("(" + re.escape(start) + ")" + re.escape(replace) + "(?=" + re.escape + ")", "\\1" + re.escape(replacement), search)

这不具备外观解决方案的优雅，但它仍然是一个非常清晰，直接的单线程。如果你看看an expert has to say on the matter（他正在谈论完全缺乏外观的JavaScript，但许多原则是相同的），你会发现他最简单的解决方案看起来很像这个。

Answer 4

我认为最好的想法就是在组中捕获您想要替换的内容，然后使用捕获的组的开始和结束属性替换它。

问候

阿德里安

#the pattern will contain the expression we want to replace as the first group
pat = "word1\s(.*)\sword2"   
test = "word1 will never be a word2"
repl = "replace"

import re
m = re.search(pat,test)

if m and m.groups() > 0:
    line = test[:m.start(1)] + repl + test[m.end(1):]
    print line
else:
    print "the pattern didn't capture any text"

这将打印： 'word1永远不会成为word2'

要替换的组可以位于字符串的任何位置。

正则表达式替换（在Python中） - 一种更简单的方法？

4 个答案: