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

Posted by Dave Griffis
Dave Griffis
2,889 Points

Can someone explain why this code doesn't pass?

Maybe I'm wrong, but I feel like the logic for this code is okay. It also produces the result I want in workspaces, but it will not pass the test in Treehouse....Any help would be appreciated. Thanks

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(topic):
    l = []
    for key, value in COURSES.items():
        if topic < value:
            l.append(key)
    return l

2 Answers

Keith Whatling
Keith Whatling
17,759 Points
Keith Whatling
Keith Whatling
17,759 Points

I find having a copy of the docs open when working with sets is a good idea, I can never remember all the operators.

The issues is where you are checking if it is in the set, you need to change the < to an &. to look for the following: s.intersection(t) or s & t = new set with elements common to s and t https://docs.python.org/2/library/sets.html

def covers(topic):
    l = []
    for key, value in COURSES.items():
        if topic & value:
            l.append(key)
    return l
Goekalp Cicek
Goekalp Cicek
9,322 Points
Goekalp Cicek
Goekalp Cicek
9,322 Points 
def covers(sett):
    l=[]
    for i in COURSES:
        if COURSES[i].intersection(sett):
            l.append(i)
    return l

def covers_all(sett):
    l=[]
    for i in COURSES:
        if sett.issubset(COURSES[i]):
            l.append(i)
    return l