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Start your free trialThurman Ward
2,718 Pointserror in task 2 is: error in task 1 ....
I cannot get past task 2. Task 1 is successful....until I "check Work" in task 2. Task too says that Task 1 is NOW not correct. I go back to task 1 - no matter what i change - it succeeds - i go to task 2 - i check work - task 1 is NOW bad again.
# 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 num_teachers(teach_dict):
the_most = 0
teachr = ''
count = 0
for key in teach_dict:
num = len(teach_dict[key])
count += 1
if the_most < num:
the_most = num
teachr = key
return count
1 Answer
Russell Sawyer
Front End Web Development Techdegree Student 15,705 PointsLeave the code from the first task as is and just add another function below it named num_teachers.
Myers Carpenter
6,421 PointsMyers Carpenter
6,421 PointsSorry that's confusing ?. Most of our code challenges build on the same piece of code, adding something each step of the way.