python changing a list into a dictionary - python-3.x

Question

so i'm taking a python class right now and am struggling with dictionaries at the moment. my assignment is simple, i have to create a fucntion "letter_positions" which will return a dictionary of all positions of a letter in a string.
for example
positions = letter_positions("fifteen e's, seven f's, four g's, six h's, eight i's, four n's, five o's, six r's, eighteen s's, eight t's, four u's, three v's, two w's, three x's")
positions['e']
should return
{4, 5, 8, 14, 16, 43, 67, 83, 88, 89, 97, 121, 122, 141, 142}
so i'm pretty much done with the assignment but i'm running into the issue that i have all values (positions) assigned to the keys (letters) as a list.
here's my code:
def letter_positions(n):
answer = {}
n = n.lower()
x = 0
for letter in n:
if letter.isalpha():
if letter not in answer:
answer[letter] = []
answer[letter].append(x)
x += 1
return answer
so instead of getting a dictionary of positions i'm getting a list of positions.
positions = letter_positions("fifteen e's, seven f's, four g's, six h's, eight i's, four n's, five o's, six r's, eighteen s's, eight t's, four u's, three v's, two w's, three x's")
positions['e']
returns
[4, 5, 8, 14, 16, 43, 67, 83, 88, 89, 97, 121, 122, 141, 142]
is there any way for me to simply change the list into a dictionary or am i approaching this in a completely wrong way?

Answer 1

change your code like this
def letter_positions(n):
answer = {}
n = n.lower()
x = 0
for letter in n:
if letter.isalpha():
answer[letter] = answer.get(letter,[])#if there is not the key letter add it as key with value an empty list
answer[letter].append(x)
x=x+1
return answer
positions = letter_positions("fifteen e's, seven f's, four g's, six h's, eight i's, four n's, five o's, six r's, eighteen s's, eight t's, four u's, three v's, two w's, three x's")
print(positions['e'])