--- slug: "replace-version-number-pattern-python" title: "Replacing Version Numbers with Python re (Regex Replacement Only Within Parentheses)" description: "Use a lambda callback with Python's `re.sub` to replace only a captured group (e.g. a version number) inside a larger pattern." url: "https://www.ytyng.com/en/blog/replace-version-number-pattern-python" publish_date: "2018-09-21T03:37:25Z" created: "2018-09-21T03:37:25Z" updated: "2026-05-11T13:11:43.202Z" categories: [] keywords: "" featured_image_url: "https://media.ytyng.com/resize/20230812/7c9a39df15584857980f3e6f95ecac5b.png.webp?width=768" has_video: false has_music: false video_urls: [] music_urls: [] lang: "en" --- # Replacing Version Numbers with Python re (Regex Replacement Only Within Parentheses)

The file looks like

version_number = "19";

and when you want to replace that file using a regular expression,

version_number = "(\d+)";

you want to use only the version number as the regular expression pattern without enclosing the other parts in pattern symbols.

However, after replacing, you want to obtain a string that includes version_number, such as

version_number = "20";

regex = re.compile('version_number = "(\d+)";')

def _replace(match):
    startindex = match.regs[1][0] - match.regs[0][0]
    end_index = match.regs[0][1] - match.regs[1][1]
    return match.group(0)[:startindex] + str(value) + match.group(0)[-end_index:]
return regex.sub(_replace, self.get_source_code())

In this way, the match indices are stored in match.regs within the _replace callback, so by slicing the string with those indices, it might work well.