Creating python scrabble function

Question

I'm trying to create a scrabble function that takes a string of letters and returns the score based on the letters. This is my code so far:

def scrabble_score(rack):
    count = 0
    for letter in rack:
        if letter == "EAIONRTLSU":
            count += 1
            return count
        elif letter == "DG":
            count += 2
            return count
        elif letter == "BCMP":
            count += 3
            return count        
        elif letter == "FHVWY":
            count += 4
            return count
        elif letter == "K":
            count += 5
            return count
        elif letter == "JX":
            count += 8
            return count
        else:
            letter == "QZ"
            count += 10
            return count

However when I tried to call the function

    scrabble_score("AABBWOL")

it returns 10 when it should return 14?

You're doing a string comparison to check if a letter exists. Use regex instead. Have a look at [this](http://stackoverflow.com/questions/5188792/how-to-check-a-string-for-specific-characters) — Dana, Apr 08 '17 at 03:03
^ expanding on @Dana, the string comparison (`==`) is returning `False` for every `if` so it gets to the last statement, the `else`, where it returns 10. — vincentmajor, Apr 08 '17 at 03:19

score 0 · Accepted Answer · edited Apr 08 '17 at 03:35

The code has two issues:

The return statement should be called only one time, after the for iteration ends, so it can sum all the letter values
each if statement should check if the letter is in the string, not if it is equal to the list of letters with a specific value.

Then it would look like:

def scrabble_score(rack):
    count = 0
    for letter in rack:
        print letter
        if letter in "EAIONRTLSU":
            count += 1
        elif letter in "DG":
            count += 2
        elif letter in "BCMP":
            count += 3
        elif letter in "FHVWY":
            count += 4
        elif letter in "K":
            count += 5
        elif letter in "JX":
            count += 8
        else:
            count += 10
    return count

returning: 14

The line: `letter in "QZ"` is not actually doing anything useful to the if-else and can be removed entirely — the_constant, Apr 08 '17 at 03:20

Creating python scrabble function

1 Answers1