Solution for average race time calculation from LinkedIn Learning Python challenge

average_race_time.py

# Source of data: https://www.arrs.run/
# This dataset has race times for women 10k runners from the Association of Road Racing Statisticians

import re
import datetime

def get_data():
    """Return content from the 10k_racetimes.txt file"""
    with open('10k_racetimes.txt', 'rt') as file:
        content = file.read()
    return content

def get_rhines_times():
    """Return a list of Jennifer Rhines' race times"""
    races = get_data()
    rhines_times = []
    def get_time(line):
        return re.findall(r'\d{2}:\S+', line)[0]

    for line in races.splitlines():
       if 'Jennifer Rhines' in line:
          rhines_times.append(get_time(line))
    return rhines_times

def get_average():
    """Return Jennifer Rhines' average race time in the format:
       mm:ss:M where :
       m corresponds to a minutes digit
       s corresponds to a seconds digit
       M corresponds to a milliseconds digit (no rounding, just the single digit)"""
    racetimes = get_rhines_times()
    total = datetime.timedelta()
    for racetime in racetimes:
        try:
            mins, secs, ms = re.split(r'[:.]', racetime)
            total += datetime.timedelta(minutes=int(mins), seconds=int(secs), milliseconds=int(ms))
        except ValueError:
            mins, secs = re.split(r'[:.]', racetime)
            total += datetime.timedelta(minutes=int(mins), seconds=int(secs))
    avg = total / len(racetimes)
    minutes = avg.seconds // 60
    seconds = avg.seconds % 60 + avg.microseconds / 1e6
    return f"{minutes:02d}:{seconds:04.1f}"

html2markdown.py

import re

ITALICS = re.compile(r'<em>(.+?)</em>')
SPACES = re.compile(r'\s+')
PARAGRAPHS = re.compile(r'<p>(.+?)</p>')
URLS = re.compile(r'<a href="(.+?)">(.+?)</a>')

def html2markdown(html):
    '''Take in html text as input and return markdown'''
    markdown = ITALICS.sub(r'*\1*', html)
    markdown = SPACES.sub(r' ', markdown)
    markdown = PARAGRAPHS.sub(r'\1\n\n', markdown)
    markdown = URLS.sub(r'[\2](\1)', markdown)
    return markdown.strip()