9/15 scrabble_score


Hi check code please?

score = {“a”: 1, “c”: 3, “b”: 3, “e”: 1, “d”: 2, “g”: 2,
“f”: 4, “i”: 1, “h”: 4, “k”: 5, “j”: 8, “m”: 3,
“l”: 1, “o”: 1, “n”: 1, “q”: 10, “p”: 3, “s”: 1,
“r”: 1, “u”: 1, “t”: 1, “w”: 4, “v”: 4, “y”: 4,
“x”: 8, “z”: 10}
def scrabble_score(word):
word = str(word)
word_as_list = list(word)
word_as_dict = {k : 0 for k in word_as_list}
word_as_key_list = []
word_as_value_list = []
for key in word_as_dict.keys():
if key in score.keys():
word_as_value_list = [score[x] for x in word_as_key_list]
personal_score = sum(word_as_value_list)
return personal_score

my error reads:
Your function fails on scrabble_score(“xenophobia”). It returns “23” when it should return “24”.


Hi @fermently,

How many times does "o" occur in “xenophobia”?
How many times can "o" occur as a key in word_as_dict?
What is the value associated with the key "o" in score?
Along with the above, note that the message reported that the score was off by 1.


Hey, before I take your message into consideration, let me thank and compliment you. This is a clear suggestion. Exactly a type of nudge required.


The above questions relate to the fact that any given key can appear only once in a dictionary.

Your scrabble_score function is more complicated than necessary. It isn’t necessary to create a new dictionary and several lists. A simpler solution could consist of the following …

  • Initialize a total to 0.
  • Iterate through the characters in word, using each individual character as a key to get its point value from the score dictionary, then adding that point value to the total.
  • After the loop, you can return the total.


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