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Susan krystof
Courses Plus Student 2,662 PointsSomeone please help with this function on sets(python)
Challenge Task 2 of 2 Great work! OK, let's create something a bit more refined. Create a new function named covers_all that takes a single set as an argument. Return the names of all of the courses, in a list, where all of the topics in the supplied set are covered. For example, covers_all({"conditions", "input"}) would return ["Python Basics", "Ruby Basics"]. Java Basics and PHP Basics would be excluded because they don't include both of those topics.
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(corses):
new_list = []
for key, value in COURSES.items():
if value.intersection(corses):
new_list.append(key)
return new_list
def covers_all(args):
new_list[]
for key, value in COURSES.items:
if args & COURSES[key] ==args:
new_list.append(key)
return new_list
1 Answer

Unsubscribed User
3,284 PointsIt seems to me that your function works, except the fact that you forgot to write some symbols: look the code below to see the difference
def covers_all(args): new_list = [] for key, value in COURSES.items(): if args & COURSES[key] ==args: new_list.append(key) return new_list
Unsubscribed User
3,284 PointsUnsubscribed User
3,284 PointsAlso, You can just go with COURSES.keys()