re.findall -> RegEx in Python

Question

import regex
frase = "text https://www.gamivo.com/product/sea-of-thieves-pc-xbox-one other text https://www.gamivo.com/product/fifa-21-origin-eng-pl-cz-tr"
x = regex.findall(r"/((http[s]?:\/\/)?(www\.)?(gamivo\.com\S*){1})", frase) 
print(x)

Result:

[('www.gamivo.com/product/sea-of-thieves-pc-xbox-one', '', 'www.', 'gamivo.com/product/sea-of-thieves-pc-xbox-one'), ('www.gamivo.com/product/fifa-21-origin-eng-pl-cz-tr', '', 'www.', 'gamivo.com/product/fifa-21-origin-eng-pl-cz-tr')]

I want something like:

[('https://www.gamivo.com/product/sea-of-thieves-pc-xbox-one', 'https://gamivo.com/product/fifa-21-origin-eng-pl-cz-tr')]

How can I do this?

Remove the first / and use non-capturing groups. r'(?:https?://)?(?:www\.)?gamivo\.com\S*', see this demo. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 23, 2021 at 9:16
do u really need regex for this ? split on spaces and take the ones with https in the resulting array — leoOrion
– leoOrion, Commented Jul 23, 2021 at 9:17
@leoOrion yes it's for a more bigger project that needs a regex. So in final project I will replace with str.replace() to use a shorted link — Diego
– Diego, Commented Jul 23, 2021 at 9:22

Wiktor Stribiżew · Accepted Answer · 2021-07-23 09:23:10Z

1

You need to

Remove the initial / char that invalidates the match of https:// / http:// since / appears after http
Remove unnecessary capturing group and {1} quantifier
Convert the optional capturing group into a non-capturing one.

See this Python demo:

import re
frase = "text https://www.gamivo.com/product/sea-of-thieves-pc-xbox-one other text https://www.gamivo.com/product/fifa-21-origin-eng-pl-cz-tr"
print( re.findall(r"(?:https?://)?(?:www\.)?gamivo\.com\S*", frase) )
# => ['https://www.gamivo.com/product/sea-of-thieves-pc-xbox-one', 'https://www.gamivo.com/product/fifa-21-origin-eng-pl-cz-tr']

See the regex demo, too. Also, see the related re.findall behaves weird post.

answered Jul 23, 2021 at 9:23

Wiktor Stribiżew

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Pausi · Accepted Answer · 2021-07-23 12:43:26Z

0

Try this, it will take string starting from https to single space or newline.

import re
frase = "text https://www.gamivo.com/product/sea-of-thieves-pc-xbox-one other text https://www.gamivo.com/product/fifa-21-origin-eng-pl-cz-tr"
x = re.findall('(https?://(?:[^\s]*))', frase)
print(x)
# ['https://www.gamivo.com/product/sea-of-thieves-pc-xbox-one', 'https://www.gamivo.com/product/fifa-21-origin-eng-pl-cz-tr']

answered Jul 23, 2021 at 12:43

Pausi

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