Python re.sub 替换为匹配的内容 [英] Python re.sub replace with matched content

问题描述

为了掌握 Python 中的正则表达式，我试图输出一些在 URL 的一部分中突出显示的 HTML.我的输入是

images/:id/size

我的输出应该是

images/:id/size

如果我在 Javascript 中执行此操作

method = 'images/:id/size';method = method.replace(/\:([a-z]+)/, '$1')警报(方法)

我得到了想要的结果，但如果我在 Python 中这样做

<预><代码>>>>方法 = '图像/:id/巨大'>>>re.sub('\:([a-z]+)', '$1', method)'图像/<span>$1</span>/huge'

我不知道，如何让 Python 返回正确的结果而不是 $1?re.sub 甚至是执行此操作的正确函数吗?

解决方案

只需使用 \1 而不是 $1:

在[1]中:导入re在 [2] 中:方法 = 'images/:id/huge'在[3]中:re.sub(r'(:[a-z]+)', r'\1', method)Out[3]: '图像/:id/huge'

另请注意原始字符串的使用 (r'...') 用于正则表达式.它不是强制性的，但不需要转义反斜杠，可以说使代码更具可读性.

Trying to get to grips with regular expressions in Python, I'm trying to output some HTML highlighted in part of a URL. My input is

images/:id/size

my output should be

images/<span>:id</span>/size

If I do this in Javascript

method = 'images/:id/size';
method = method.replace(/\:([a-z]+)/, '<span>$1</span>')
alert(method)

I get the desired result, but if I do this in Python

>>> method = 'images/:id/huge'
>>> re.sub('\:([a-z]+)', '<span>$1</span>', method)
'images/<span>$1</span>/huge'

I don't, how do I get Python to return the correct result rather than $1? Is re.sub even the right function to do this?

解决方案

Simply use \1 instead of $1:

In [1]: import re

In [2]: method = 'images/:id/huge'

In [3]: re.sub(r'(:[a-z]+)', r'<span>\1</span>', method)
Out[3]: 'images/<span>:id</span>/huge'

Also note the use of raw strings (r'...') for regular expressions. It is not mandatory but removes the need to escape backslashes, arguably making the code slightly more readable.

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