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Rahmat Abdhir
15,198 PointsHow to return teacher with most_count for my most_classes() dictionary
def most_classes(teachers): my_dict = {} max_count = 0; for teacher in teachers: my_dict[teacher] = len(teachers[teacher]) #return my_dict
{'Jason Seifer': 3, 'Jon Low': 5, 'Kenneth Love': 2}
for each_key in my_dict.keys():
if my_dict[each_key] > max_count:
max_count = my_dict[each_key]
#now given the dict value how you get the key
return max_count
d = {'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'], 'Kenneth Love': ['Python Basics', 'Python Collections'], 'Jon Low': ['python', 'ruby', 'angular', 'javascript', 'java'] }
print(most_classes(d))
# 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):
my_dict = {}
max_count = 0;
for key in dicts:
my_dict[key] = len(dicts[key])
#return my_dict
#{'Jason Seifer': 3, 'Jon Low': 5, 'Kenneth Love': 2}
for each_key in my_dict.keys():
if my_dict[each_key] > max_count:
max_count = my_dict[each_key]
return each_key
1 Answer
![Russell Sawyer](https://uploads.teamtreehouse.com/production/profile-photos/6176732/micro_profile-trees.jpg)
Russell Sawyer
Front End Web Development Techdegree Student 15,705 PointsYou're pretty close.
def most_classes(dicts):
most_class = ""
max_count = 0
for teacher in dicts:
if len(dicts[teacher]) > max_count:
max_count = len(dicts[teacher])
most_class = teacher
return most_class