是否有一个匹配和替换功能？

Question

我想知道是否有更简单的替代方法（例如单个函数调用）来匹配和替换以下示例：

>>> import re
>>> 
>>> line = 'file:///windows-d/academic%20discipline/study%20objects/areas/formal%20systems/math'
>>> 
>>> match = re.match(r'^file://(.*)$', line)
>>> if match and match.group(1):
...     substitution = re.sub(r'%20', r' ', match.group(1))
... 
>>> substitution
'/windows-d/academic discipline/study objects/areas/formal systems/math'

感谢。

Answer 1

我要回避你的正则表达式问题并建议你使用别的东西：

>>> line = 'file:///windows-d/academic%20discipline/study%20objects/areas/formal%20systems/math' 
>>> import urllib
>>> urllib.unquote(line)
'file:///windows-d/academic discipline/study objects/areas/formal systems/math'

然后用切片剥离file://或必要时str.replace。

%20（空格）不是此处唯一可能的转义字符，所以最好使用正确的工具进行工作，而不是让你的正则表达式解决方案在以后有另一个字符需要时逃跑。

Answer 2

您可以尝试以下简单的python代码，

>>> import re
>>> line = 'file:///windows-d/academic%20discipline/study%20objects/areas/formal%20systems/math'
>>> m = re.sub(r'%20|file://', r' ', line).strip()
>>> m
'/windows-d/academic discipline/study objects/areas/formal systems/math'

re.sub(r'%20|file://', r' ', line).strip()代码用空格替换字符串%20或file://。同样，strip()函数会删除所有前导和尾随空格。

Answer 3

>>> import re
>>> s = 'file:///windows-d/academic%20discipline/study%20objects/areas/formal%20systems/math'
>>> re.sub(r'^file://(.*)$', lambda m: m.group(1).replace('%20', ' '), s)
'/windows-d/academic discipline/study objects/areas/formal systems/math'

---

>>> s = 'file:///windows-d/academic%20discipline/study%20objects/areas/formal%20systems/math'
>>> s.replace('file://', '').replace('%20', ' ')
'/windows-d/academic discipline/study objects/areas/formal systems/math'