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Python Python Collections (2016, retired 2019) Sets Set Math

Posted by Kirome Thompson
Kirome Thompson
5,350 Points

Help writing a covers_all function, that takes a single argument and returns the names of all the courses in a list

Great work! OK, let's create something a bit more refined. Create a new function named covers_all that takes a single set as an argument. Return the names of all of the courses, in a list, where all of the topics in the supplied set are covered. For example, covers_all({"conditions", "input"}) would return ["Python Basics", "Ruby Basics"]. Java Basics and PHP Basics would be excluded because they don't include both of those topics.

sets.py 
COURSES = {
    "Python Basics": {"Python", "functions", "variables",
                      "booleans", "integers", "floats",
                      "arrays", "strings", "exceptions",
                      "conditions", "input", "loops"},
    "Java Basics": {"Java", "strings", "variables",
                    "input", "exceptions", "integers",
                    "booleans", "loops"},
    "PHP Basics": {"PHP", "variables", "conditions",
                   "integers", "floats", "strings",
                   "booleans", "HTML"},
    "Ruby Basics": {"Ruby", "strings", "floats",
                    "integers", "conditions",
                    "functions", "input"}
}

def covers(topics):
    course_list = []
    for key, value in COURSES.items():
        if topics.intersection(value):
            course_list.append(key)
    return course_list

def covers_all(all_topics):
    course_list = []
    for key, value in COURSES.items():
        if all_topics.union(value):
            course_list.append(key)
    # print('course list is: {}'.format(course_list))
    return course_list

1 Answer

Mustafa Başaran
Mustafa Başaran
28,046 Points
Mustafa Başaran
Mustafa Başaran
28,046 Points

Hello Kirome,

The second step of the challenge asks you to find courses that include both topics. You can check this condition through issubset(other).. that tests whether every element in the set is in other. So, please drop union and try issubset instead. I hope this helps.

Kirome Thompson
Kirome Thompson
5,350 Points

Hi Mustafa

Yes issubset() worked a charm thanks for the help I appreciate it.