not able to get the mistake in the code

Question

def most_classes(dict): dict = {'PQR':['ruby foundation', 'ruby on rails', 'Tech foundations'], 'ABC':['python basics' , 'python collection']} max_count = 0 for key in dict: if value in key: count[value] +=1 print(max(count[value]), dict[key])

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(dict):
  dict = {'PQR':['ruby foundation', 'ruby on rails', 'Tech foundations'], 'ABC':['python basics' , 'python collection']}
  max_count = 0
  for key in dict:
    if value in key:
      count[value] +=1
  print(max(count[value]), dict[key])

Answer 1 · 2015-04-11T21:26:04Z

April 11, 2015 9:26pm

There a few things to clear up in your code.

dict does not need to be defined. It's value is set by the arguments when most_classes() is referenced by the grader code. Comment out dict = .... line
The count dictionary needs to be initialized. Add count = {} to init as empty dictionary.
When checking if value is present in the count dict, you need to reference the value as dict[key] and to check against keys count.keys(). That is: if dict[key] in count.keys()
You need to compare the current count[value] against max_count to see if a new maximum has been reached.
As the hints suggest "Better hold onto the teacher name somewhere", perhaps in a new variable max_teacher
While a print shows the values, you need to use a return statement to pass values back to the grader. Replace print() with return. And return the teacher's name ("max_teacher") and the count value ("count[value]"). Note return does not need parens.
Best to use 4-space indent instead of 2-spaces.

def most_classes(dict):
    dict = {'PQR':['ruby foundation', 'ruby on rails', 'Tech foundations'], 'ABC':['python basics' , 'python collection']}
    max_count = 0
    for key in dict:
        if value in key:
          count[value] +=1
    print(max(count[value]), dict[key])

Answer 2 · 2015-04-11T20:28:23Z

April 11, 2015 8:28pm

This is the code I got for this challenge:

# 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(teacher_dict):
  max_count = 0
  busy_teacher = ''
  for teacher1,teacher2 in teacher_dict.items():
      if len(teacher2) > max_count:
        max_count = len(teacher2)
        busy_teacher = teacher1
  return busy_teacher

def num_teachers(dct):
  return len(dct)

def stats(arg3):
  teacherList = []
  for key in arg3:
    tempList = []
    tempList.append(key)
    tempList.append(len(arg3[key]))
    teacherList.append(tempList)
  return teacherList

def courses(teachers):
  total_courses = []
  end_result = []
  for key in teachers.items():
    total_courses.append(teachers.values())
  for courses_of_one_teacher in teachers.values():
    for course in courses_of_one_teacher:
      end_result.append(course)
  return end_result

Answer 3 · 2015-04-11T21:36:58Z

April 11, 2015 9:36pm

Thanks Chris!

Welcome to the Treehouse Community

Looking to learn something new?

Arun Shivaramakrishna

Arun Shivaramakrishna

not able to get the mistake in the code

3 Answers

Chris Freeman

Chris Freeman

Devin Scheu

Devin Scheu

Arun Shivaramakrishna

Arun Shivaramakrishna