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

Alex Avila
Alex Avila
8,570 Points

What is a simpler way to do the teacher stats code challenge?

I resisted the urge to google the answers and this is what I came up with. However, I feel like there's a much simpler way to do these functions. Especially the most_courses function.

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(d):
    return len(d.keys())


def num_courses(d):
    total_courses = []
    for value in d.values():
        for val in value:
            total_courses.append(val)
    return len(total_courses)


def courses(d):
    total_courses = []
    for value in d.values():
        for val in value:
            total_courses.append(val)
    return total_courses

def most_courses(d):
    courses_num_list = []
    teachers_list = []
    indexes = []
    all_indexes = []
    most_teachers = []

    for courses in d.values():
        courses_num = len(courses)
        courses_num_list.append(courses_num)
    for teachers in d.keys():
        teachers_list.append(teachers)

    count = 0
    for num in courses_num_list:
        if count == 0:
            high_val = num
            count += 1
        else:
            if num > high_val:
                high_val = num
            else:
                pass

    indexes = [index for index, value in enumerate(courses_num_list) if value == high_val]
    for i in indexes:
        index = i
        most_teachers.append(teachers_list[index])

    for i in most_teachers:
        return i

def stats(d):
    teacher = []
    courses = []
    list_of_lists = []
    for key in d.keys():
        teacher.append(key)
    for value in d.values():
        courses.append(len(value))

    for num in range(0, len(teacher)):
        list_within = [teacher[num], courses[num]]
        list_of_lists.append(list_within)

    return list_of_lists

2 Answers

Steven Parker
Steven Parker
231,007 Points

You sure have a heck of a function there, but .. it worked! :+1:

But you definitely were working way to hard. Your first 3 functions were pretty concise .. I wonder why this one threw you?

But anyway, I don't normally give spoilers, but I'll show you how I did just that one:

def most_courses(d):
    num = 0
    for teacher in d:
        if len(d[teacher]) > num:
            num =  len(d[teacher])
            most = teacher
    return most
Alex Avila
Alex Avila
8,570 Points

I think I didn't know I could do something like 'most = teacher', then 'return most'. I would've guessed that would give me a key value pair from the dict. Something like '{teacher: [course1, course2, etc.]}'.

But, that is SO much simpler. Thanks!

Steven Parker
Steven Parker
231,007 Points

When you iterate on a dictionary, you get the keys. If you want the pair you can iterate on dict.items() instead

I just had a run through and came up with this (spoiler warning) https://pastebin.com/UceQr596

<<code removed to paste bin, so you don't accidentally see solution code >>

In methodology a lot of it insn't that much different, it just uses built in method calls. That makes it a bit easier to read, and they tend to be programmed in C rather than Python so run a bit faster. This is in no way a definitive list - I'm sure there are much more efficient ones, this is just a balance of what I know, what's easy to read, etc.

Ah I haddn't thought about people stumbling on to it by accident.. moved it to pastebin