Python 中的拼字游戏查找器

Question

Implement the word_calculator method to return the correct scrabble word score. The scores are already set up to be use and are managed within dictionaries:

two = ['d', 'g']
three = ['b', 'c', 'm', 'p']
four = ['f', 'h', 'v', 'w', 'y']
five = ['k']
eight = ['j', 'x']
ten = ['q', 'z']

with having these list, i need to produce a function def word_calculator(word): where i need to pass a string to the method. the parameter zoo should return 12 , and bus should return 5 .

Can anyone help me to produce the function please?

Answer 1

This, like many problems in computer science, is much easier to do if you use a better data-structure.

As you have it, you will have to loop over all six lists for each letter, check if the letter is in each of them, and then hard code the score for each list programmatically so that you can add it to some total. That's a lot of work, and a lot of messy code.

However, you could use a dictionary instead. A dictionary lets you map any value to any other value. So, for example

x = {"A": 1, "B": 3}

Means that x["A"] is 1, and that x["B"] is 3, etc.

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Lets imagine we had a dictionary containing the mapping of every letter to its scrabble value:

scores = {"A": 1, "B": 3, "C": 3, "D": 2, "E": 1, "F": 4, "G": 2, "H": 4, "I": 1, "J": 8, "K": 5, "L": 1, "M": 3, "N": 1, "O": 1, "P": 3, "Q": 1, "R": 1, "S": 1, "T": 1, "U": 1, "V": 4, "W": 4, "X": 8, "Y": 4, "Z": 10}

How would we find the word score for "ZOO" ? We would need to:

1. Loop through every letter
2. Find the score for the letter (eg scores[letter] )
3. Add the score to some running total

For example,

total = 0
for letter in "ZOO" :
    total = total + score[letter]
print(total) # Prints 12

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If you want to be fancy, this can also be done all on the one line using a list comprehension:

sum([scores[letter] for letter in "ZOO"])

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Further reading:

• https://docs.python.org/3/tutorial/datastructures.html

  (explains dictionaries very well, and also talks about list comprehensions)

Answer 2

# the dictionary of scores to letter lists
score_letters = {
    2: ['d', 'g'],
    3: ['b', 'c', 'm', 'p'],
    4: ['f', 'h', 'v', 'w', 'y'],
    5: ['k'],
    8: ['j', 'x'],
    10: ['q', 'z']
}

# use `score_letters` to make a list from letters to score values
letter_scores = {letter: score for score, letters in score_letters.items() for letter in letters}
# equivalent to:
# letter_scores={}
# for score, letters in score_letters.items():
#     for letter in letters:
#         letter_scores[letter] = score

print('letter_scores =', letter_scores)


def word_calculator(word):
    # sum the scores of each letter in the word
    # `letter_scores.get(letter, 1)` means
    #     "get the value associated with the
    #     key `letter`, or if there is no such
    #     key in the dictionary, use the
    #     default value 1"
    return sum(letter_scores.get(letter, 1) for letter in word)


print('zoo', word_calculator('zoo'))
print('bus', word_calculator('bus'))