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Start your free trialKudakwashe Chatikobo
1,010 Pointsneed help am failing to get the required courses
.
# The dictionary will look something like:
# {'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'],
# 'Kenneth Love': ['Python Basics', 'Python Collections']}
#
# Each key will be a Teacher and the value will be a list of courses.
#
# Your code goes below here.
def num_teachers(teachers):
return int(len(teachers))
def num_courses(tt):
courses = 0
for value in tt.values():
courses += len(value)
return courses
def courses(slist):
for courses in slist.values():
course = list(courses)
return course
1 Answer
Antonio De Rose
20,885 PointsNot sure how that works, but I was able to get around that issue by doing the below
# the reason for me to have 2 for loops, was to have one flat list, if I stop from the first loop
# and then to append, I would have had a list of lists
#think of it like, the moment you have tt.values(), you result in something lie
# tt = [
# ['jQuery Basics', 'Node.js Basics'],
# ['Python Basics', 'Python Collections']
# ]
# when you loop for the first time, you have a list of lists
# value = ['jQuery Basics', 'Node.js Basics'],
# the above itself will not work, as it is another list
# cause, we are interested in the strings. will have to break till you have a string in hand
# hence loop over again, for the tx to have 'jQuery Basics', 'Node.js Basics'
# append it to courses, the newly created list, for the return
# it will come out of the inner loop, as it had finished the inner loop
# now go to the outer loop, have got the next one remaining, ['Python Basics', 'Python Collections']
# do same
Dict = {
'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'],
'Kenneth Love': ['Python Basics', 'Python Collections']
}
def num_courses(tt):
courses = []
for value in tt.values():
for tx in value:
courses.append(tx)
return courses
print(num_courses(Dict))