I have a table with urls like
vk.com/albums54751623?z=photo54751623_341094858%2Fphotos54751623
vk.com/albums54751623
vk.com/id36375649
vk.com/id36375649
I need to find all urls like vk.com/id36375649 (only id)
I try
for url in urls:
if url == re.compile('vk.com/^[a-z0-9]'):
print url
else:
continue
but this is uncorrectly, because it didn't return anything
You can use startswith:
strs = ['vk.com/albums54751623?z=photo54751623_341094858%2Fphotos54751623',
'vk.com/albums54751623',
'vk.com/id36375649',
'vk.com/id36375649']
print([x for x in strs if x.startswith(r'vk.com/id')])
See the IDEONE demo
UPDATE
To address the issues stated in comments below this answer, you will have to use a regex with some checks:
^vk\.com/(?!album)\w+$
See the regex demo and a Python demo:
import re
strs = ['vk.com/albums54751623?z=photo54751623_341094858%2Fphotos54751623',
'vk.com/albums54751623',
'vk.com/id36375649',
'vk.com/id36375649',
'vk.com/id36375649?z=album-28413960_228518010',
'vk.com/tania_sevostianova'
]
print([x for x in strs if re.search(r'^vk\.com/(?!album)\w+$', x)])
# => ['vk.com/id36375649', 'vk.com/id36375649', 'vk.com/tania_sevostianova']
r'vk\.com/id\d+' can be very handyr'^vk\.com/(?!album)\w+$'. See the update.A regular expression like the following might work
vk.com\/id\d+
Remember that in regex you need to escape certain characters like slashes.
