If structure

Yacht
Yacht in Python

The constants here can be set to random, null, or numeric values, and an if structure inside the score function can determine the code to be executed.

As one-liners aren't necessary here, we can spread out the code to make it look neater:

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

Note that the code inside the if statements themselves can differ, but the key idea here is to use if and elif to branch out the code, and return 0 at the end if nothing else has been returned. The if condition itself can be different, with people commonly checking if category == ONES as opposed to category == 'ONES' (or whatever the dummy value is).

This may not be an ideal way to solve the exercise, as the code is rather long and convoluted. However, it is a valid (and fast) solution. Using structural pattern matching, introduced in Python 3.10, could shorten and clarify the code in this situation. Pulling some logic out of the score function and into additional "helper" functions could also help.

13th Nov 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):
    match category:
        case 1 | 2 | 3 | 4 | 5 | 6:
            return dice.count(category) * category
        case _:
            return 0
Structural Pattern Matching

Use structural pattern matching