Not really sure how to complete this challenge. Please help!

...I can try but please be aware I'm still a newbie.

It really helped me to look at the python docs he recommends in 'Set Math'. I played around a bit in the shell to get the hang of the different methods and also watched that video a few times.

def covers(arg):
    hold = []  
    for key, value in COURSES.items():
        #key is the course names (e.g. Python, Ruby) and value is the sets{'Python', 'functions' etc.} that we want to compare
        if value & arg:
        #so value(set1).intersection(set2) - finds common items between sets
            hold.append(key)
    return hold
covers({'Ruby'})

As for the second challenge, you'll need to check out the method .issubset().

I can post the answer if you're struggling but I'm sure you'll get it once you get the hang of the common set operators (|, &, ^ and -) and have practised with the subset method in the shell.

Don't give up - it took me hours to get there! Hope this is helpful!

disregard, finally figured it out...not sure I fully understand, but this passed the test:

def covers_all(single_set):
    new_list = []
    for key, value in COURSES.items():
        if set(single_set).issubset(value) :
            new_list.append(key)
    return new_list

An easier way to understand (We are outputting answer like a list item because the answer should be like " ['Python Basics'] "

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

Here is a solution using set operators instead of lists.

def covers(arg):

solution = []

for key, value in COURSES.items():

    if set(arg) & set(value):

        solution.append(key)

return(solution)

hey @chriskellett, I sure could use the help on task 2...am stuuuuccckkk - thx

def covers(a): b = [] for k,v in COURSES.items(): if v & a: b.append(k) return b

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

I used this code and it seems to work.

I did a bunch of tests in workspaces. The problem is the code works only if the argument is given in the same format as the string it has to search for in the courses dictionary's set. For eg. inputing {"strings"} will return ['Python Basics', 'Java Basics', PHP Basics', 'Ruby Basics'], or inputting {"Python"} will return ['Python Basics'].

Lets say we input {"Strings"} instead of {"strings"} or {"python'} instead of {"Python"}, the code returns an empty list: []. Its because the format of the string in argument doesn't match the format of the course in the dictionary while it runs the loop, and thus doesn't append to the empty list we created.

What is the work around for this so the string format doesn't matter?

def covers(my_topics):
    courses = []
    for course, course_topics in COURSES.items():
        if course_topics.intersection(my_topics): #equivalent to course_topics & my_topics
           courses.append(course)
    return courses

Here's the answer with better naming conventions so that it's easier to understand what is going on.

Here's how I did it, folks!

def covers(topic_set):
    course_list = []
    for course, topics in COURSES.items():
        if topics & topic_set:
            course_list.append(course)
    return course_list

def covers_all(topic_set):
    course_list = []
    for course, topics in COURSES.items():
        if not topic_set - topics:
            course_list.append(course)
    return course_list