Python Collections: teachers_stats challenge, Task 1 of 4. I passed, but wonder if there is a cleaner alternative?

Question

Hi there! Just throwing this question out there. I completed task 1 of 4 in the teachers_stats challenge, and my code was successful! However, it seems like my logic is rather complicated. I am wondering if anyone has a more simple alternative? Attaching my code below for discussion. Thanks!!

teachers.py

# The dictionary will be something like:
# {'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'],
#  'Kenneth Love': ['Python Basics', 'Python Collections']}
#
# Often, it's a good idea to hold onto a max_count variable.
# Update it when you find a teacher with more classes than
# the current count. Better hold onto the teacher name somewhere
# too!
#
# Your code goes below here.

def most_classes(teacherdict):
    newdict = {}
    for key, value in teacherdict.items():
        newdict[len(value)]= key
    most = list(sorted(newdict.keys()))
    return newdict[most[-1]]

Answer 1 · 2015-09-20T20:40:15Z

September 20, 2015 8:40pm

You need to return the teacher who teaches the most classes. What you do is create a dictionary from the number of classes to teacher. You sort the dictionary on the keys, turn it into a list and grab and return the teacher that is last in the list. It is certainly a solution.

A simpler solution is to create a pair (teacher, numberOfClasses). Then for each iteration on the input dictionary you update that pair to contain the teacher with the highest number of classes of the current "best" teacher and the one examined in the iteration. Does that make sense? Then when you're done with the loop you'll know that the teacher in your pair is the right one and so you can return it.

Answer 2 · 2015-09-23T15:22:32Z

September 23, 2015 3:22pm

I have done this task before but never used the sorted function. Although since your are converting the dictionary values into keys and in turn the keys into a list you could use the max function to achieve the result with even less code.

def most_classes(teacherdict):
    newdict = {}
    for key, value in teacherdict.items():
        newdict[len(value)]= key
    return newdict[max(newdict.keys())]

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lindseyk

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Python Collections: teachers_stats challenge, Task 1 of 4. I passed, but wonder if there is a cleaner alternative?

2 Answers

Emil Rais

Emil Rais

lindseyk

lindseyk

Andreas cormack

Andreas cormack

lindseyk

lindseyk