Solved Exercise: Validate and Extract Data with Regex in Python

The solution separates two responsibilities. The function parse_line receives an already stripped line and returns a tuple with the fields if it is valid, or None if it is not. The main program walks through the input, accumulates results and formats the output.

The regex captures named groups and is compiled once at module level. The check letter is verified outside the regex, because it depends on an arithmetic computation on the number.

import re
import sys

LETTERS = "TRWAGMYFPDXBNJZSQVHLCKE"
NAME_CHARS = "A-Za-zÁÉÍÓÚÜÑáéíóúüñ"

PATTERN = re.compile(
    r"(?P<num>\d{8})(?P<letter>[A-Z]);"
    r"(?P<surname>[" + NAME_CHARS + r"][" + NAME_CHARS + r" -]*), "
    r"(?P<name>[" + NAME_CHARS + r"][" + NAME_CHARS + r" -]*);"
    r"(?P<course>[A-Z]{3}-\d{3});"
    r"(?P<mark>10(?:\.0{1,2})?|\d(?:\.\d{1,2})?)"
)


def parse_line(line):
    m = PATTERN.fullmatch(line)
    if m is None:
        return None
    if LETTERS[int(m.group("num")) % 23] != m.group("letter"):
        return None
    student_id = m.group("num") + m.group("letter")
    return student_id, m.group("surname"), m.group("name"), m.group("course"), float(m.group("mark"))


def main():
    marks = {}
    rejected = []
    valid = 0
    best = None
    for number, raw in enumerate(sys.stdin.read().splitlines(), 1):
        line = raw.strip()
        if not line:
            continue
        data = parse_line(line)
        if data is None:
            rejected.append(number)
            continue
        valid += 1
        student_id, surname, name, course, mark = data
        marks.setdefault(course, []).append(mark)
        if best is None or mark > best[3]:
            best = (student_id, name, surname, mark)
    print(f"VALID: {valid}")
    print(f"REJECTED: {len(rejected)}")
    for k in rejected:
        print(f"REJECTED LINE {k}")
    for course in sorted(marks):
        values = marks[course]
        mean = sum(values) / len(values)
        print(f"{course}: students={len(values)} mean={mean:.2f}")
    if best is not None:
        student_id, name, surname, mark = best
        print(f"BEST: {student_id} {name} {surname} {mark:.2f}")


main()

The constant PATTERN is built by concatenating fragments so that it stays readable. The groups num and letter separate the eight digits from the letter, so that parse_line can compute int(m.group("num")) % 23 and compare it with LETTERS. The groups surname and name must start with a letter and allow spaces and hyphens; the comma and the space between them are literal parts of the pattern, which is why Ruiz,Ana is rejected.

The group mark uses two alternatives: 10(?:\.0{1,2})? for ten (with optional zero decimals) and \d(?:\.\d{1,2})? for a digit from 0 to 9 with up to two decimals. That rules out 10.5, 7.123 and 07. Because we use fullmatch, no anchors are needed and the classic error of accepting a partial match disappears.

In main, enumerate(..., 1) numbers every line before the blank ones are filtered, as the statement requires. Valid lines feed the dictionary marks (course to list of marks) and the variable best, which is only updated with mark > best[3]: the strict comparison guarantees that on a tie the first student in the input is kept. The output walks sorted(marks) for alphabetical order and formats the mean with :.2f.

In the statement's example test there are six useful lines: three valid and three rejected (wrong ID letter, mark 11, and a missing space after the comma); line 3 is blank and just takes up a number. In the test with only blank lines the program prints just the two counters at zero, and in the tie test it checks that BEST keeps the first student.