Structural Pattern Matching

Yacht
Yacht in Python

Another very interesting approach is to use structural pattern matching. Existing in Python since 3.10, this feature allows for neater code than traditional if structures.

By and large, we reuse the code from the if structure approach. We set the constants to random values and check for them in the match structure. category is the "subject", and in every other line, we check it against a "pattern".

ONES = 1
TWOS = 2
THREES = 3
FOURS = 4
FIVES = 5
SIXES = 6
FULL_HOUSE = 'FULL_HOUSE'
FOUR_OF_A_KIND = 'FOUR_OF_A_KIND'
LITTLE_STRAIGHT = 'LITTLE_STRAIGHT'
BIG_STRAIGHT = 'BIG_STRAIGHT'
CHOICE = 'CHOICE'
YACHT = 'YACHT'

def score(dice, category):
    match category:
        case 1 | 2 | 3 | 4 | 5 | 6:
            return dice.count(category) * category
        case 'FULL_HOUSE' if len(set(dice)) == 2 and dice.count(dice[0]) in [2, 3]:
            return sum(dice)
        case 'FOUR_OF_A_KIND' if dice[0] == dice[3] or dice[1] == dice[4]:
            return dice[1] * 4
        case 'LITTLE_STRAIGHT' if sorted(dice) == [1, 2, 3, 4, 5]:
            return 30
        case 'BIG_STRAIGHT' if sorted(dice) == [2, 3, 4, 5, 6]:
            return 30 
        case 'YACHT' if all(num == dice[0] for num in dice):
            return 50
        case 'CHOICE':
            return sum(dice)
        case _:
            return 0

For the first pattern, we utilize "or patterns", using the | operator. This checks whether the subject is any of the provided patterns.

In the next five patterns, we check an additional condition along with the pattern matching. Finally, we use the wildcard operator _ to match anything. As the compiler checks the patterns (cases) in order, return 0 will be executed if none of the other patterns match.

Note that the conditions might differ, but the patterns must have hard coded values - that is, you can't say case ONES ... instead of case 1 .... This will capture the category and lead to unexpected behavior.

This code is much clenaer than the corresponding if structure code.

8th May 2024 · Found it useful?

Other Approaches to Yacht in Python

Other ways our community solved this exercise
def digits(num):
    return lambda dice: dice.count(num) * num
YACHT = lambda x: 50 if x.count(x[0]) == len(x) else 0
ONES = digits(1)
FULL_HOUSE = lambda x: sum(x) if len(set(x)) == 2 and x.count(x[0]) in [2, 3] else 0
LITTLE_STRAIGHT = lambda x: 30 if sorted(x) == [1, 2, 3, 4, 5] else 0
def score(dice, category):
    return category(dice)
Lambdas with Functions

Use lambdas with functions 

ONES = 1
YACHT = 'YACHT'
def score(dice, category):
    if category == 'ONES':
        ...
    elif category == 'FULL_HOUSE':
        ...
    return 0
If structure

Use an if structure