Check for funny characters with a regexp

[bertilow]

I want to be able to check if a string contains any of the following "funny" characters:

"ā㥹ĆćČčĎďĐđēĕĖėęěĞğġīįİıĽľŁłŃńņňŋŌōőŒœŕřŚśŞşŠšŢţťŧũūŮűųŹźŻżŽžſ"

Let's say the string is "abcdefghijklmnopqrstuvwxyz".

This is just a small element of a large application where for complicated reasons the check must be done with a regular expression and sometimes in a case-insentitive way. The check can be just a part of a larger regexp that can contain practially anything else.

I tried the following obvious solution:

re.findall('(?i)[ā㥹ĆćČčĎďĐđēĕĖėęěĞğġīįİıĽľŁłŃńņňŋŌōőŒœŕřŚśŞşŠšŢţťŧũūŮűųŹźŻżŽžſ]', 'abcdefghijklmnopqrstuvwxyz')

Result:

['i', 's']

That is a fail. The characters "i" and "s" are not in that regular expression. If i remove "(?i)" it works, but that is not an option.

Is there any way to solve this? (The same thing works without any problem if I use Perl...)

I have tried everything I can think of. Any advice is welcome.

I use Python 3 (of course).