Third section of code isn't passing

Question

Ok guys... Where have I gone wrong in the Stats function... (sigh)

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.
most_classes = {'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'],
                'Kenneth Love': ['Python Basics', 'Python Collections']}
def most_classes(my_dict):
  teachers_list = {}
  for teacher in my_dict:
    teachers_list[teacher] = len(my_dict[teacher])
  return max(teachers_list, key=teachers_list.get)

def num_teachers(teachers):
  teacherNumber = 0
  for teacher in teachers:
    teacherNumber += 1
  return teacherNumber

def stats(teachers):
  newlist = []
  for name, item in teachers:
    newlist.append((teachers[name], len(teachers[item]))
  return newlist

Answer 1 · 2015-08-06T06:58:15Z

August 6, 2015 6:58am

Hi, Carina, This line

for name, item in teachers:

is not valid Python code, you can't have more than 1 iterating variable while doing for loop. Additionally, on the following line

    newlist.append((teachers[name], len(teachers[item]))

There're 3 opening parenthesis VS 2 closing parenthesis, this is a syntax error; also, elements enclosed with () is tuple in Python, but this part of challenge is asking for list within list, so you need to use [] literal instead.

One way to fix that is

def stats(teachers):
  newlist = []
  for name in teachers:
    newlist.append([name, len(teachers[name])])
  return newlist

Alternatively, if you really want to retrieve both the key and value each time you loop through the dictionary, you may use the .items() method on the dictionary.

def stats(teachers):
  newlist = []
  for key, value in teachers.items():
    newlist.append([key, len(value)])  # append [<name>, <number of classes>] to newlist
  return newlist

Answer 2 · 2015-08-06T07:21:00Z

August 6, 2015 7:21am

The problem you're having is that he's requesting a nested list. So it should output something like: [ ["kenneth love", 5], ["jason seifer", 3] ]. You were trying to create a list of names and numbers stored in tuples. However, I don't believe that you can nest the append method that way, either. Here's how I solved the problem:

def stats(dic):
    big_list = []
    for key in dic:
        little_list = []
        little_list.append(key)
        little_list.append( len(dic[key]) )
        big_list.append(little_list)
    return big_list

For every entry in the dictionary, I created a new list and appended the name and, then, the number of classes. Lastly, I appended the smaller list to the larger one. You could probably condense this a little but I wanted it to be easy to read.

Answer 3 · 2015-08-06T14:17:35Z

August 6, 2015 2:17pm

Got this one working (but stuck on last question):

def stats(my_dict): my_list = [] for key in my_dict: num_classes = len(my_dict[key]) my_list.append([key,num_classes]) return my_list

The number of classes is simply the length of the values associated with each key. Then append the key with the length value to a temp list and return the result. I think i could combine the two lines in the for loop but I kept them separate to make it easier to read and understand. Hope this helps.

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Carina De Jager

Carina De Jager

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