尝试从多个选项中选择 select，然后添加正确的选项 - 非常困惑

Question

if (dice1 == dice2) or (dice1 == dice3) or (dice2 == dice1) or (dice2 == dice3) 
or (dice3 == dice2) or (dice3 == dice1):
     
    score = # no clue what to put here

I need to find a way to do; if any of these options are correct then the correct option will be chosen and the dice variables are added together, eg Are two or more dice equal? YES, then score = sum of two equal dice .

Really confused, can anyone offer help?

Answer 1

It's not necessary to ask dice2 == dice1 if you know the answer to dice1 == dice2 . And could be a good idea to set score to 0 even if none of the dice are equal

With just 3 dice it's probably easiest and fastest to just test the 3 possible pairs.

if dice1 == dice2 or dice1 == dice3:
    score = dice1 * 2
elif dice2 == dice3:
    score = dice2 * 2
else:
    score = 0

If you have more than 3 dice you need to make more decisions of which score to give if there are more than one pair of equals. Here's a fast solution that returns the score from the highest pair:

def best_pair(dice):
    candidates = set()
    best = 0
    for die in dice:
        # don't bother if it's not better than best
        if die > best:
            # did we find an equal before?
            if die in candidates:
                # then it's the new best
                best = die
            else:
                # maybe next time...
                candidates.add(die)
    return best * 2

best_pair([1, 3, 1, 4, 3, 5, 3, 6, 1, 2, 4, 5, 1, 1, 1])
# 10

The collections.Counter.most_common() suggested in answer by @kaya is a possibility too but if the dice are (1, 1, 5, 5) the first item of most_common() will be the pair of ones. This is probably not expected if the dice is thrown at once. A solution to this could be to find the maximum from collections.Counter.items() instead:

from collections import Counter
dice = [1, 3, 1, 4, 3, 5, 3, 6, 1, 2, 4, 5, 1, 1, 1]
score = 2 * max((v for v, c in Counter(dice).items() if c>=2), default = 0)
# 10

Answer 2

This is probably overkill just for three dice, but here is one way to do it that doesn't require duplicating code: the collections.Counter class has a most_common method which can tell you both what value occurs most commonly, and how many times it occurs.

from collections import Counter

def score(*dice):
    (value, count), = Counter(dice).most_common(1)
    return 2 * value if count >= 2 else 0

Examples:

>>> score(6, 3, 6)
12
>>> score(4, 2, 3)
0
>>> score(5, 5, 5)
10