other way to write this task?

Question

a passed this task by writing the code below,but the #instructions told me to use the max_count variable, so i think there might be another way to write the code, but i have no idea, someone help me about these?

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(dicts):
    dicts={'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'],
         'Kenneth Love': ['Python Basics', 'Python Collections']}
    lists={}
    for key in dicts:
        lists.update({key:len(dicts[key])})
    for teacher in lists:
        if lists[teacher] == (max(lists.values())):
            return(teacher)

Answer 1 · 2016-12-04T11:21:04Z

December 4, 2016 11:21am

Hi there.

Here the alternative solution, using only 1 for loop, the code is pretty easy to read, I think you should have no problem understanding it.

def most_classes(dic):
    teacher = ""                        # name of the teacher w/ most classes
    most_class = 0                      # numbers of the most classes
    for key in dic:
        if len(dic[key]) > most_class:  # if this teacher has more classes
            most_class = len(dic[key])  # re-assign most_class variable
            teacher = key               # re-assign teacher's name
    return teacher

Later when you finish the Python functional programming course, it'll be possible to have a simpler and more effecient solution by using the lambda expression like this.

def most_classes(dic):
    return max(dic, key=lambda i: len(dic[i]))

Hope it helps

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other way to write this task?

1 Answer

William Li

William Li

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