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Python Python Collections (2016, retired 2019) Dictionaries Teacher Stats

Kars Jansens
Kars Jansens
5,348 Points

I don't know why this isn't working {python}

question: Wow, I just can't stump you! OK, two more to go. I think this one's my favorite, though. Create a function named most_courses that takes our good ol' teacher dictionary. most_courses should return the name of the teacher with the most courses. You might need to hold onto some sort of max count variable.

Can someone help me?

teachers.py
# 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(strs):
    count = 0
    for teacher in strs.keys():
        count += 1
    return count
def num_courses(strs):
    count = 0
    for value in strs.values():
        for item in value:
            count += 1
    return count

def courses(strs):
    lista = []
    for value in strs.values():
        for item in value:
            lista.append(item)
    return lista
def most_courses(strs):
    dicta = dict()
    lista = []
    for item in strs.items():
        count = 0
        x, y = item
        for course in y:
            count +=1
        dicta[count] = x
    for key in dicta.keys():
        lista.append(key)
    KEY = max(lista)
    teacher = strs[KEY]
    return teacher

1 Answer

Steven Parker
Steven Parker
221,070 Points

Your method is unusual and a bit complicated — but essentially sound. And you are very close to having it working!

Once you've identified the largest number of courses in "KEY", you need the teacher associated with that course. But the code is trying to look up the count in the original dictionary instead of the one you built with counts as keys and teachers as values ("dicta"):

    teacher = strs[KEY]   # instead of using the original dictionary...
    teacher = dicta[KEY]  # look up the count in your custom one instead