Great work! OK, let's create something a bit more refined. Create a new function named covers_all that takes a single s

Question

i am lost in this one

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(arg):
    list_of_courses= []
    for key, value in COURSES.items():
        if value & arg:
            list_of_courses.append(key)
    return list_of_courses

def covers_all(arg):
    list_of_all_courses = []
    for course in COURSES.items():
        if arg & COURSES [course] == arg:
            list_of_all_courses.append(course)
    return list_of_all_courses

Answer 1 · 2018-03-06T19:31:15Z

March 6, 2018 7:31pm

H Mamadou,

The solution I used for the second part checks that both sets have the same values.

it's quite similar to the first task but also checking to see if the intersection set is identical to the argument set

def covers_all(arg):
    course_list = []
    for key, value in COURSES.items():
        if (arg & value) == arg:
            course_list.append(key)
    return course_list

Hope this helps

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mamadou diene

mamadou diene

Great work! OK, let's create something a bit more refined. Create a new function named covers_all that takes a single s

1 Answer

Kenichi Beveridge

Kenichi Beveridge