Exercise 2 solution

    import re
    from collections import Counter

    tw = re.compile(r"(?P<INT>\d+)|(?P<WORD>[a-z]+)|(?P<SP>\s+)|(?P<X>.)")
    s = "go 12 steps, then 7 more!"

    print([(m.lastgroup, m.group()) for m in tw.finditer(s) if m.lastgroup != "SP"])
    # [('WORD', 'go'), ('INT', '12'), ('WORD', 'steps'), ('X', ','), ('WORD', 'then'), ('INT', '7'), ('WORD', 'more'), ('X', '!')]

    print(Counter(m.lastgroup for m in tw.finditer(s)))
    # Counter({'SP': 5, 'WORD': 4, 'INT': 2, 'X': 2})

(All lines were run and printed exactly what is shown.)

Explanation: each alternative is its own named group, so m.lastgroup is the
kind of token. The catch-all X comes last so that it only matches characters
no earlier alternative wanted. The spaces are filtered out of the list but are
still counted in the Counter, because that counts every match.