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yannis sousanis
7,199 Pointscant believe that most of every step of this challenge I need help
It works and I tested on my ide but not here.
# The dictionary will look something like:
# {'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'],
# 'Kenneth Love': ['Python Basics', 'Python Collections']}
#
# Each key will be a Teacher and the value will be a list of courses.
#
# Your code goes below here.
def num_teachers(a):
return len(a)
def num_courses(dic):
number = 0
for v in dic.values():
number += len(v)
return number
def courses(dic):
course_list = []
for v in dic.values():
course_list.append(v)
return course_list
def courses(dic):
course_list = []
for v in dic.values():
for a in v:
course_list.append(a)
return course_list
def most_courses(dic)
m_courses = 0
for k , v in dic.items():
if len(v) > m_courses :
m_courses = len(v)
teacher = k
return teacher
2 Answers

Steven Parker
216,057 PointsIt looks like you have a couple of typos on the definition line of most_courses:
- the line is indented by 1 space
- there's no colon at the end of the line
Did it really work like that in your IDE?

yannis sousanis
7,199 Pointsthank you ...these typos are ruining my life, I wrote it correct on my ide but not here, I understand that the colon is important but I dont understand why does one space doesnt run the code

Steven Parker
216,057 PointsI'm not sure which, but if it accepts it as an indent, that means function would be defined inside the previous one and not available in the global scope. And otherwise, it would be an inconsistent indentation error.
Jordan Hoover
Python Web Development Techdegree Graduate 59,267 PointsJordan Hoover
Python Web Development Techdegree Graduate 59,267 PointsAnd this, I though that was me.