How can create a regex class that is the intersection of two other regex classes? For example, how can I search for consonants with the [a-z] and [^aeiou] without explicitly constructing a regex class containing all the consonants like so:
[bcdfghjlkmnpqrstvwxyz] # explicit consonant regex class
This regex should do the trick : (?=[^aeiou])(?=[a-z]).
The first group (?=...) asserts that the pattern [^aeiou] can be matched, then restarts the matching at the beginning and moves on to the second pattern (which works the same way), it's like a logical AND, and the whole regex will only match if all of these two expressions match.
As an alternative to Python's re module, you can do this explicitly with the regex library, which supports set operations for character classes:
The operators, in order of increasing precedence, are:
||for union (“x||y”means “x or y”)
~~(double tilde) for symmetric difference (“x~~y”means “x or y, but not > both”)
&&for intersection (“x&&y”means “x and y”)
--(double dash) for difference (“x––y”means “x but not y”)
So to match only consonants, your regular expression could be:
>>> regex.findall('[[a-z]&&[^aeiou]]+', 'abcde', regex.VERSION1)
['bcd']
Or equivalently using set difference:
>>> regex.findall('[[a-z]--[aeiou]]+', 'abcde', regex.VERSION1)
['bcd']
The character class difference or intersection is not available in the re module, so what you can do?
using ranges:
[bcdfghj-np-tv-z]
using the \w character class:
[^\W0-9_aeiouAEIOU]
a lookahead (not very efficient since you need to make a test for each character):
(?:(?![eiou])[b-z])
using the new regex module that has the difference feature:
[[b-z]--[eiou]]
