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1,714 PointsWorks fine in console ???
def num_teachers(kwarg): teachers={kwarg} count=0 for key in teachers.keys(): count+=1 return count
# 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(**kwarg):
teachers={**kwarg}
count=0
for key in teachers.keys():
count+=1
return count
1 Answer
Chris Freeman
Treehouse Moderator 68,454 PointsDon't include the "**" in the parameter list. You don't need to use it at all. Replacing "**kwargs" with "teachers" also works.
In fact, there is no difference:
$ python
Python 3.5.3 (v3.5.3:1880cb95a742, Jan 16 2017, 16:02:32) [MSC v.1900 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> dict1 = {'Andrew Chalkley': ['jQuery Basics', 'Node.js Basics'], 'Kenneth Love': ['Python Basics', 'Python Collectio
ns']}
>>> dict2 = {**dict1}
>>> dict1 == dict2
True