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I have a list of urls which have varying numbers in the end after an & sign. I'm not able to apply a regex to remove these numbers (including the &) from the url as there are multiple & in the string and the re.sub('&\d*',"",x) command filters all of the & including the one I want to remove.

The url is: http://helloworld.com?p1=123&p2=987&hello=world&123456

The desired output I want is: http://helloworld.com?p1=123&p2=987&hello=world

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  • Also as a side note I think you mean to be using forward slashes in your urls like this http:// not back slashes. Commented Nov 15, 2014 at 8:55

2 Answers 2

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You can use an anchored pattern if you always want the last parameter:

re.sub(r'&\d+$',"",x)

The important piece is the dollar sign which says to only match at the end.

Also you should keep in mind that whenever you use * that can match the empty string. If you want to match a non-empty string, you need to use +.

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@Kasra your answer is wrong because you still may match something in the middle of the string.
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You need + after \d for match! When you use &\d*, it first matches the middle &. Also, you need a $ to specify that your pattern is at the end of the string:

'http:\\helloworld.com?p1=123&p2=987&hello=world&123456'
                             ^

so use re.sub(r'(&\d+)$',"",x) instead of yours! See DEMO.

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In general it is best to use raw strings when specifying regexes using r''

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