Python骰子游戏使用列表

Question

我正在尝试开发一个从1开始的骰子值序列运行的python骰子游戏，从1开始的值是3分。 如果没有1，则该手为0，而一个骰子的手数可以超过一个。

例如[5,6,2,4,2,3,6,4]为0

[3,4,1,5,3,1,4,6]是6（两个1）

[2,1,1,1,2,3,3,2]为24（两个1,2,3 = 18，一个1,2 = 6）

[5,3,2,6,4,5,1,4]是18（一个1,2,3,4,5,6）

def get_hand_score(list_of_dice):
   score1 = 0
   list1_sorted = list_of_dice.sort()
   if "1" in str(list_of_dice):
      score1+=3
      if "2" in str(list_of_dice):
        score1+=3
        if "3" in str(list_of_dice):
          score1+=3
          if "4" in str(list_of_dice):
             score1+=3
             if "5" in str(list_of_dice):
               score1+=3
               if "6" in str(list_of_dice):
                 score1+=3

return score1


def test_get_hand_score():
   print("1.  score: ", get_hand_score([5, 3, 2, 5, 5, 6, 4, 3]))
   print("2.  score: ", get_hand_score([3, 4, 1, 5, 3, 1, 4, 6]))
   print("3.  score: ", get_hand_score([5, 3, 2, 6, 4, 5, 1, 4]))
   print("4.  score: ", get_hand_score([2, 1, 1, 1, 2, 3, 3, 2])) 
   print("5.  score: ", get_hand_score([3, 4, 1, 5, 2, 1, 4, 6]))

   list1 = [5, 3, 2, 5, 5, 6, 4, 3]
   print("6.  dice: ", list1)
   score1 = get_hand_score(list1)   
   list1.sort()
   print("    dice_sorted: ", list1)
   print("    score:", score1)
   print()

   list1 = [5, 3, 2, 6, 4, 5, 1, 4]
   print("7.  dice: ", list1)
   list1_sorted = sorted(list1)
   score1 = get_hand_score(list1)
   print("    dice_sorted: ", list1_sorted)
   print("    score:", score1)
   print()

   list1 = [2, 1, 1, 1, 2, 3, 3, 2]     
   print("8.  dice: ", list1)
   list1_sorted = sorted(list1)
   score1 = get_hand_score(list1)
   print("    dice_sorted: ", list1_sorted)
   print("    score:", score1)
   print()

这给了我：

1. score:  0
2. score:  3
3. score:  18
4. score:  9
5. score:  18
6.  dice:  [5, 3, 2, 5, 5, 6, 4, 3]
    dice_sorted:  [2, 3, 3, 4, 5, 5, 5, 6]
    score: 0

7.  dice:  [5, 3, 2, 6, 4, 5, 1, 4]
    dice_sorted:  [1, 2, 3, 4, 4, 5, 5, 6]
    score: 18

8.  dice:  [2, 1, 1, 1, 2, 3, 3, 2]
    dice_sorted:  [1, 1, 1, 2, 2, 2, 3, 3]
    score: 9

预期结果：

1. score:  0
2. score:  6
3. score:  18
4. score:  24
5. score:  21
6.  dice:  [5, 3, 2, 5, 5, 6, 4, 3]
    dice_sorted:  [2, 3, 3, 4, 5, 5, 5, 6]
    score: 0

7.  dice:  [5, 3, 2, 6, 4, 5, 1, 4]
    dice_sorted:  [1, 2, 3, 4, 4, 5, 5, 6]
    score: 18

8.  dice:  [2, 1, 1, 1, 2, 3, 3, 2]
    dice_sorted:  [1, 1, 1, 2, 2, 2, 3, 3]
    score: 24

Answer 1

此代码有效（产生您的预期输出）：

from collections import Counter

def get_hand_score(dice):
    counts = Counter(dice)

    total = 0

    while counts[1] > 0:
        n = 1
        while counts[n] > 0:
            counts[n] -= 1
            total += 1
            n += 1

    return total * 3

代码的问题在于，您只能算一个序列。 您先寻找1，然后寻找2，依此类推，但是您不会以任何方式将其标记为“二手”，然后重复。 因此，当您的代码看到123123时，它只计数一次123。

编辑

更简单的（也许是？）工作代码：

def get_hand_score(dice):
    counts = Counter(dice)

    n = counts[1]
    score = 0

    for i in range(1, 7):
        n = min(n, counts[i])
        score += n

    return score * 3

这里的想法是，每找到1分，得分就会提高。 （至少，它只是一个序列的一部分。）

之后，每2到1的分数都会增加分数。 并且每3到2的总数，分数就会再次增加。