Dictionary

Scrabble Score
Scrabble Score in Python
LETTER_SCORES = {
    'A': 1, 'E': 1, 'I': 1, 'O': 1, 'U': 1,
    'L': 1, 'N': 1, 'R': 1, 'S': 1, 'T': 1,
    'D': 2, 'G': 2, 'B': 3, 'C': 3, 'M': 3,
    'P': 3, 'F': 4, 'H': 4, 'V': 4, 'W': 4,
    'Y': 4, 'K': 5, 'J': 8, 'X': 8, 'Q': 10, 'Z': 10
}
def score(word):
    return sum(LETTER_SCORES[letter] for letter in word.upper())

This code starts with defining a constant LETTER_SCORES as a dictionary (dicts) where each letter is a key and the corresponding score is a value. Then the score function is defined, which takes a <word> as an argument.

The function returns the total score for the word using the built-in function sum. Sum is passed a generator expression that iterates over the letters in the word, looking up each score in LETTER_SCORES. The generator expression produces the score values on the fly. This means that it doesn't use memory to store all the values from LETTER_SCORES. Instead, each value is looked up as needed by sum.

Within the generator expression, the word is converted from lower to uppercase. Each letter of the word is looked up in LETTER_SCORES, and the score value is yielded to sum as sum iterates over the expression. This is almost exactly the same process as using a list comprehension. However, a list comprehension would look up the values and save them into a list in memory. sum would then "unpack" or iterate over the list.

A variation on this dictionary approach is to use a dictionary transposition.

LETTER_SCORES = {
    1: {'A', 'E', 'I', 'O', 'U', 'L', 'N', 'R', 'S', 'T'},
    2: {'D', 'G'},
    3: {'B', 'C', 'M', 'P'},
    4: {'F', 'H', 'V', 'W', 'Y'},
    5: {'K'},
    8: {'J', 'X'},
    10: {'Q', 'Z'}
}

def score(word):
    return sum(next(score for score, letters in LETTER_SCORES.items() if character in letters) for character in word.upper())

However, transposing the dictionary so that the keys are the score and the values are the letters requires more computational calculation (a loop within a loop) and is harder to read. Therefore, arranging the dictionary by letter is both more efficient and easier to understand.

29th May 2024 · Found it useful?