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

Posted by Marco Koopman
Marco Koopman
5,290 Points

I got the right output but in the wrong way

This is a massive workaround but how can I make this code smaller? I know this is not the right way to do it.

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(param):

    my_list = list(param)
    my_list.sort()

    for course in COURSES:

      for c in COURSES[course]:

        if(my_list[0] == c):

          return [course]

1 Answer

Grigorij Schleifer
Grigorij Schleifer
10,365 Points
Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hi Marco, try this one:

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

It uses the intersection shortcut and appends the intersection result to a list.