Exercise 3 solution

    import re
    urls = "http://a.com/x https://b.org/y/z ftp://c.net"

    print(re.findall(r"(https?)://([\w.]+)", urls))
    # [('http', 'a.com'), ('https', 'b.org')]
    print(re.findall(r"(?:https?)://([\w.]+)", urls))
    # ['a.com', 'b.org']
    print(re.findall(r"(?P<proto>https?)://(?P<host>[\w.]+)", urls))
    # [('http', 'a.com'), ('https', 'b.org')]
    for m in re.finditer(r"(?P<proto>https?)://(?P<host>[\w.]+)", urls):
        print(m.groupdict())
    # {'proto': 'http', 'host': 'a.com'}
    # {'proto': 'https', 'host': 'b.org'}

(All lines were run and printed exactly the output shown.)

Explanation: findall returns tuples when there are two or more capturing
groups, and plain strings when there is exactly one. Making the protocol
group non-capturing leaves a single group, so only the host is returned.
Naming the groups does not change what findall returns, but finditer with
groupdict() gives each match as a labelled dictionary. The ftp address is
skipped in every case because the pattern only allows http or https.