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# stuck at teacher stats 3 of 4

Maybe im just slow but these teacher stats exercises are pretty hard.

what is the solution to this?

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(my_dict):
max_count = 0
teacher = ""                       # teacher variable holds the name of teacher with most courses
for i in my_dict:
if len(my_dict[i]) > max_count:  # if the current teacher has more courses than max_count
max_count = len(my_dict[i])    # reassign the max_count
teacher = i                    # reassign the name of this teacher to teacher variable
return teacher

def num_teachers(my_dict):
count = 0
for i in my_dict:
count += 1
return count

def stats(my_dict):
new_word = []
for key in my_dict:
new_words = "{}, {}".format(key, len(my_dict[key]))
return new_words.append(new_word)
``` Oh sorry i did not look at that..u don't need that statement, so it should be:

```def stats(my_dict):
teacher_list = []
for i in my_dict:
count = len(my_dict[i])
teacher = i
teacher_list.append([teacher,count])
return teacher_list
```

i was wondering what it was for I was like. whaaaaat thank you very much ```def stats(my_dict):
teacher_list = []
for i in my_dict:
if len(my_dict[i]):
count = len(my_dict[i])
teacher = i
teacher_list.append([teacher,count])
return teacher_list
```

i don't get the " if len(my_dict[i]):" what does this do? That solution seems a bit overly complicated.... try this:

```def stats(n):
tStats = []
z = n.items()
for x, y in z:
tStats.append([x, len(y)])
return tStats
```