consecutive punctuation and alpha-numeric characters

Question

Probably a simple question, but I don't have a lot of regex experience. I would like to take a string and select all the consecutive punctuation characters and all the consecutive alpha-numeric characters

This is as close as I could get

r="my9zza :)asax"
import re
re.findall(r'(\w+)|([^a-zA-Z0-9\s]+)', r)

returns

[('my9zza', ''), ('', ':)'), ('asax', '')]

but I would like

['my9zza', ':)', 'asax']

Answer 1

Simply use:

r = "my9zza :)asax"
import re
print(re.findall(r'\w+|[^a-zA-Z0-9\s]+', r))

The problem was having two sets of parentheses in your original code, causing findall to return a 2-ple.

If you wanted to keep the original regex, you can also easily transform your result into the desired output with:

[x[0] or x[1] for x in result]

Answer 2

You can try this:

s = [('my9zza', ''), ('', ':)'), ('asax', '')]
final_s = [[b for b in i if b][0] for i in s]

Output:

['my9zza', ':)', 'asax']