I feel stuck. Can someone offer a hand?

Question

I've really tried making this work on my laptop but I can't seem to find a solution.

I know I need to: 1) See how long each key value is 2)Somehow compare them 3)Return the key of the longer key value

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(dic):
    return len(dic.keys())

def num_courses(dic):
    sum = 0
    for key_value in dic:
        sum += len(dic[key_value])
    return sum

def courses(dic):
    new_list = []
    for key_value in dic:
        new_list.extend(dic[key_value])

    return new_list    


def most_courses(dic):
    for key in dic.keys():
    x = len(dic[keys])

Answer 1 · 2017-01-19T15:56:45Z

January 19, 2017 3:56pm

Let's look at your objectives individually:

1) See how long each ~~key~~ value is

Simple enough, you did this already in num_courses, using the len function.

2) Somehow compare them

You could use an if statement with a "greater than" (>) comparison operator between the two values.

3) Return the key of the longer key value

In your loop, you could keep both the longest length and the name (key) that goes with it. Then, when the loop finishes you just return that retained name.

Answer 2 · 2017-01-20T09:59:12Z

January 20, 2017 9:59am

Simple logic:

def most_courses(dic):

     max_len = 0
     teacher_with_most_courses = ''
     for key in dic.keys():
         if len(dic[key]) >= max_len: 
             max_len = len(dic[key])
             teacher_with_most_courses = key
     return teacher_with_most_courses

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Rosu Calin

Rosu Calin

I feel stuck. Can someone offer a hand?

2 Answers

Steven Parker

Steven Parker

Let's look at your objectives individually:

Ronit Mankad

Ronit Mankad