正则表达式删除单词和查找单词

Question

I would like to find and delete the https part in a sentence. I use re.search("^https://t.co/.*[a-zA-Z]" ,data)` and the result is:

match='https://xx.x/ekGSeJufuH 7 jalan indonesia yang pa

match='https://xx.x/okbymT3g'

But I want to just take match='https://xx.x/ekGSeJufuH and delete while keeping the rest of the word. I there something wrong with my regex?

Answer 1

.* matches any characters including whitespace.

An easier way is that

1. find a sentense starting with 'https://',
2. find the first whitespace(' ') in the sentence,
3. delete substring before the whitespace.

I think it works because the URL doesn't allow any WS inside.

Answer 2

From what I understand, you just want to exclude the "https://" from the string. If so, this may be a regular expression you're looking for:

r"https://(.*)"

Using the above regular expression with the addresses you provided:

>>> regex = re.compile(r"https://(.*)")
>>> regex.search("https://xx.x/ekGSeJufuH 7 jalan indonesia yang pa").group(1)
'xx.x/ekGSeJufuH 7 jalan indonesia yang pa'
>>> regex.search("https://xx.x/okbymT3g").group(1)
'xx.x/okbymT3g'

If there are more criteria for the regular expression which I missed, just comment on my answer and I'll update the regular expression accordingly.

Answer 3

I tried to tweak it a little bit and solves it with

re.search("^https://t.co/\S*",txt)