python中的正则表达式:是否有可能获得匹配、替换和最终字符串? [英] Regex in python: is it possible to get the match, replacement, and final string?

问题描述

要进行正则表达式替换，您需要提供三样东西:

For doing a regex substitution, there are three things that you give it:

• 匹配模式
• 替换模式
• 原始字符串

正则表达式引擎发现我感兴趣的三件事:

There are three things that the regex engine finds that are of interest to me:

• 匹配的字符串
• 替换字符串
• 最终处理的字符串

当使用 re.sub 时，最终的字符串是返回的内容.但是是否可以访问其他两个东西，匹配字符串和替换字符串?

When using re.sub, the final string is what's returned. But is it possible to access the other two things, the matched string and replacement string?

这是一个例子:

orig = "This is the original string."
matchpat = "(orig.*?l)"
replacepat = "not the \\1"

final = re.sub(matchpat, replacepat, orig)
print(final)
# This is the not the original string

匹配字符串是"original"，替换字符串是"not the original".有没有办法得到它们?我正在编写一个脚本来搜索和替换许多文件，我希望它打印它正在查找和替换的内容，而不打印整行.

The match string is "original" and the replacement string is "not the original". Is there a way to get them? I'm writing a script to to search and replace in many files, and I want it to print it what it's finding and replacing, without printing out the entire line.

推荐答案

class Replacement(object):

    def __init__(self, replacement):
        self.replacement = replacement
        self.matched = None
        self.replaced = None

    def __call__(self, match):
        self.matched = match.group(0)
        self.replaced = match.expand(self.replacement)
        return self.replaced

>>> repl = Replacement('not the \\1')
>>> re.sub('(orig.*?l)', repl, 'This is the original string.')
    'This is the not the original string.'
>>> repl.matched
    'original'
>>> repl.replaced
    'not the original'

<小时>

正如@F.J 指出的那样，上面只会记住最后一次匹配/替换.此版本处理多次出现:

---

as @F.J has pointed out, the above will remember only the last match/replacement. This version handles multiple occurrences:

class Replacement(object):

    def __init__(self, replacement):
        self.replacement = replacement
        self.occurrences = []

    def __call__(self, match):
        matched = match.group(0)
        replaced = match.expand(self.replacement)
        self.occurrences.append((matched, replaced))
        return replaced

>>> repl = Replacement('[\\1]')
>>> re.sub('\s(\d)', repl, '1 2 3')
    '1[2][3]'

>>> for matched, replaced in repl.occurrences:
   ....:     print matched, '=>', replaced
   ....:     
 2 => [2]
 3 => [3]

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