Python - 拼字游戏字数

Question

I need to write a short program for scrabble where given a word it can calculate the score.

I am new to python and still figuring out how to combine lists. Any help or an explanation to the solution would be much appreciated.

I think i have to state that each list is equal to a number and then using that work out how to look at a word and calculate that score.

Would anyone be able to give me a solution and talk me through the various points?

one_letter_point = ['e', 'a', 'o', 't', 'i', 'n', 'r', 's', 'l', 'u'] 
two_letter_point = ['d', 'g']
three_letter_point = ['c', 'm', 'b', 'p']
four_letter_point = ['h', 'f', 'w', 'y', 'v']
five_letter_point = ['k']
eight_letter_point = ['j', 'x']
ten_letter_point = ['q', 'z']

def scrabble_word_count(word):

    one_letter_word = 1 #do the same for the rest and then not sure what to do

    return scrabble_word_count(word) # not sure if this is supposed to be done 

print answer_word('zoo')

I have this alternate method, but it only looks at the first letter and gives me the point for that. eg points for apple = 1. as it only looks at the a.

def scrabble_word_count(word):
   score = 0
   for letter in word:
    if letter in one_letter_point:
        score += 1
    elif letter in two_letter_point:
        score += 2
    elif letter in three_letter_point:
         score += 3
    elif letter in four_letter_point:
         score += 4
    elif letter in five_letter_point:
          score += 5
    elif letter in eight_letter_point:
          score += 8
    elif letter in ten_letter_point:
          score += 10
    return score

    print(scrabble_word_count('apple'))

Answer 1

You need to specify the correspondance between letters and value, the better to get quick access is to have a simple correspondance for each letter.

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}

Then for each letter of the word, find it in the score dict and sum all

# simple for loop style
def scrabble_word_count(word):
    res = 0
    for letter in word:
        res += score[letter]
    return res

# functionnal style
def scrabble_word_count(word):
    return sum(map(score.get, word))

---

Using separate lists :

one_letter_point = ['e', 'a', 'o', 't', 'i', 'n', 'r', 's', 'l', 'u']
two_letter_point = ['d', 'g']
three_letter_point = ['c', 'm', 'b', 'p']
four_letter_point = ['h', 'f', 'w', 'y', 'p']
five_letter_point = ['k']
eight_letter_point = ['j']
ten_letter_point = ['q', 'z']

def scrabble_word_count(word):
    res = 0
    for letter in word:
        if letter in one_letter_point:
            res += 1
        elif letter in two_letter_point:
            res += 2
        elif letter in three_letter_point:
            res += 3
        elif letter in four_letter_point:
            res += 4
        elif letter in five_letter_point:
            res += 5
        elif letter in eight_letter_point:
            res += 8
        elif letter in ten_letter_point:
            res += 10
    return res

Answer 2

Good start, but your data structure creates a problem. Okay, so with five_letter_point = ['k'] , we know that a 'k' is worth five points. How can we extract that data from the fact that it's stored under a variable name containing the word "five"? We really can't.

Let's use a dict instead, which is also known outside the context of Python as a hashmap. For example with secret_identities = {"Batman": "Bruce Wayne", "Kal-El": "Superman"} you can look up secret_identities['Batman'] and learn that Batman is Bruce Wayne. Or in your case:

letter_scores = {
    'e': 1,
    'a': 1,
    ...
    'd': 2,
    ...
}

Now the scores are stored as a dict, which maps each letter to the number of points that it gets. For instance, you can look the score for 'e' up with letter_scores['e'] , which should return the number 1 .

Answer 3

Use multiple if statements to add to the appropriate value to the score depending on which list the letter is in.

def scabble_word_count(word):
    score = 0
    for letter in word:
        if letter in one_letter_point:
            score += 1
        elif letter in two_letter_point:
            score += 2
        elif letter in three_letter_point:
            score += 3
        ...
    return score

You can avoid all the repeated code by putting all the letter lists into another list. Use empty lists for the points that don't have any letters.

all_letters = [
    one_letter_point, two_letter_point, three_letter_point, four_letter_point, five_letter_point, 
    [], [], [], eight_letter_point, [], ten_letter_point
]
def scrabble_word_count(word):
    score = 0
    for letter in word:
        for index, letters in enumerate(all_letters):
            if letter in letters:
                score += index + 1 # indexes start at 0, points start at 1
                break
    return score