intersection vs isdisjoint for sets math code challenge "write a function named covers"

Question

# this code does not work in the code challenge but does work in Anacoda.

def covers(set_of_topics):
    list_of_course = []
    for key in COURSES:
        if COURSES[key].intersection(set_of_topics) == set_of_topics:
            list_of_course.append(key)
    return list_of_course

#this codes works in Anaconda and passes the code challenge-- WHY?

def covers(set_of_topics):
    courses_from_course = []
    for key in COURSES:
        if COURSES[key].isdisjoint(set_of_topics) == False:
            courses_from_course.append(key)
        else:
            continue
    return courses_from_course

Answer 1 · 2017-12-23T23:10:19Z

December 23, 2017 11:10pm

Your first covers function returns courses where all of the items in the passed in set are present. Meaning that if I passed in: {"Python", "strings"} for example then I would get back ['Python Basics']. Since that is the only course that contains both a "Python" section and a "strings" section.

Your second covers function returns courses where any of the items in the passed in set are present. Meaning that if I passed in: {"Python", "strings"} for example then I would get back ['Python Basics', 'Java Basics', 'PHP Basics', 'Ruby Basics']. Since all of those course contain either a "Python" section or a "string" section.

For the first task Treehouse wants a function that behaves like your second cover function, giving back any course that contains any of the items passed in. For the second task where you are asked to create a covers_all function Treehouse actually looks for a function that behaves like your first covers function.

Which means that if you rename your first covers function to covers_all then you can complete the entire challenge with the two functions you have written.

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Hector Perez

Hector Perez

intersection vs isdisjoint for sets math code challenge "write a function named covers"

1 Answer

andren

andren