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10,503 Points(Stats Challenge) Code Check: Room for Improvement?
# 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(dict_teachers):
trophyTeacher = ""
mostClasses = 0
numClasses = 0
for teacher in dict_teachers:
numClasses = len(dict_teachers[teacher])
if numClasses > mostClasses:
mostClasses = numClasses
trophyTeacher = teacher
return trophyTeacher
def num_teachers(dict_teachers):
amountTeachers = 0
for teacher in dict_teachers:
amountTeachers += 1
return amountTeachers
def stats(dict_teachers):
new_list = []
second_list = []
numClasses = 0
for teacher in dict_teachers:
numClasses = len(dict_teachers[teacher])
new_list = teacher.split('<string to list>')
new_list.append(numClasses)
second_list.append(new_list)
return second_list
def courses(dict_teachers):
new_list = []
for courses in dict_teachers.values():
for course in courses:
new_list.append(course)
return new_list
1 Answer
Chris Freeman
Treehouse Moderator 68,457 PointsGood job on the challenge. Your code works as is. I would give the following simplifications and some more Pythonic ways to code. See embedded comments:
# The dictionary will be something like:
# {'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'],
# 'Kenneth Love': ['Python Basics', 'Python Collections']}
# Recommended by PEP-8 styling
# Use 4-space indentation instead of 2 spaces
# Use underscore instead of camelCase for variable names. For example:
# num_classes instead of numClasses
# most_classes instead of mostClasses
def most_classes(dict_teachers):
trophyTeacher = ""
mostClasses = 0
# numClasses = 0 #<-- initialization not needed since it is set in each loop pass
# alternative for loop structure using enurmerate
# for teacher, class_list in enumerate(dict_teachers):
for teacher in dict_teachers:
# num_classes = len(class_list)
numClasses = len(dict_teachers[teacher])
if numClasses > mostClasses:
mostClasses = numClasses
trophyTeacher = teacher
return trophyTeacher
def num_teachers(dict_teachers):
# This entire function can be simplified to
# return len(dict_teachers)
amountTeachers = 0
for teacher in dict_teachers:
amountTeachers += 1
return amountTeachers
def stats(dict_teachers):
# new_list = [] #<-- initialization not needed
second_list = []
# numClasses = 0 #<-- initialization not needed
# enumerate could also be used here
for teacher in dict_teachers:
numClasses = len(dict_teachers[teacher])
# Since the split string is not present in 'teacher', no split will happen
# new_list = teacher.split('<string to list>')
# instead simply start the new_list with 'teacher'
new_list = [teacher]
new_list.append(numClasses)
second_list.append(new_list)
return second_list
def courses(dict_teachers):
new_list = []
for courses in dict_teachers.values():
# instead of looping through each course in courses, used the 'extend()' method
# for course in courses:
# new_list.append(course)
new_list.extend(courses)
return new_list
This is a first pass review. Feel free to post your updated code for more review. Keep up the good work!
qlpxjevhuv
10,503 Pointsqlpxjevhuv
10,503 PointsAh, I can see I made a bunch of silly mistakes! Thank you for your evaluation!