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Start your free trialMinseok Kim
2,708 PointsDon't know how to do the sets challenge...
https://teamtreehouse.com/library/python-collections-2/sets/set-math
Challenge: Let's write some functions to explore set math a bit more. We're going to be using this COURSES dict in all of the examples. Don't change it, though!
So, first, write a function named covers that accepts a single parameter, a set of topics. Have the function return a list of courses from COURSES where the supplied set and the course's value (also a set) overlap.
For example, covers({"Python"}) would return ["Python Basics"].
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(a):
list_1 = []
for value in a:
for courses in COURSES:
if courses & value:
list_1.append(courses)
return list_1
Minseok Kim
2,708 PointsSam Wederell
then why doesnt this work?
def covers(a):
list_1 = []
for courses in COURSES:
if courses & a:
list_1.append(courses)
print (list_1)
Sam Wederell
20,276 Pointsyou're comparing the course objects rather than the values.
1 Answer
Steven Parker
231,269 PointsYou're close, in your 2nd try, but there's two issues:
- you need to test the associated set instead of the key ("
if COURSES[courses] & a:
") - you need to return the new list (and you won't need to "print" anything)
Sam Wederell
20,276 PointsSam Wederell
20,276 PointsYou don't need to iterate through the supplied argument 'a'.