Python计算掷骰子数和数倍数

Question

问题是：我需要掷3个骰子。如果两个（或三个）骰子返回相同的数字，停止。如果3个骰子都是唯一的（例如2、4和6），则再次掷骰子。进行此操作，直到两倍/三倍滚动，或7次，以先到者为准。在

注意：我是一个python新手。在

到目前为止，我所得到的只是生成216种可能的组合：

import itertools

all_possible = list(itertools.permutations([1,2,3,4,5,6],3))
input = raw_input()

print all_possible

生成这种类型的输出：

[(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 2), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 2), (1, 4, 3), (1, 4, 5), (1, 4, 6), (1, 5, 2), (1, 5, 3), (1, 5, 4), (1, 5, 6), (1, 6, 2), (1, 6, 3), (1, 6, 4), (1, 6, 5), (2, 1, 3), (2, 1, 4), (2, 1, 5), (2, 1, 6), (2, 3, 1), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 1), (2, 4, 3), (2, 4, 5), (2, 4, 6), (2, 5, 1), (2, 5, 3), (2, 5, 4), (2, 5, 6), (2, 6, 1), (2, 6, 3), (2, 6, 4), (2, 6, 5), (3, 1, 2), (3, 1, 4), (3, 1, 5), (3, 1, 6), (3, 2, 1), (3, 2, 4), (3, 2, 5), (3, 2, 6), (3, 4, 1), (3, 4, 2), (3, 4, 5), (3, 4, 6), (3, 5, 1), (3, 5, 2), (3, 5, 4), (3, 5, 6), (3, 6, 1), (3, 6, 2), (3, 6, 4), (3, 6, 5), (4, 1, 2), (4, 1, 3), (4, 1, 5), (4, 1, 6), (4, 2, 1), (4, 2, 3), (4, 2, 5), (4, 2, 6), (4, 3, 1), (4, 3, 2), (4, 3, 5), (4, 3, 6), (4, 5, 1), (4, 5, 2), (4, 5, 3), (4, 5, 6), (4, 6, 1), (4, 6, 2), (4, 6, 3), (4, 6, 5), (5, 1, 2), (5, 1, 3), (5, 1, 4), (5, 1, 6), (5, 2, 1), (5, 2, 3), (5, 2, 4), (5, 2, 6), (5, 3, 1), (5, 3, 2), (5, 3, 4), (5, 3, 6), (5, 4, 1), (5, 4, 2), (5, 4, 3), (5, 4, 6), (5, 6, 1), (5, 6, 2), (5, 6, 3), (5, 6, 4), (6, 1, 2), (6, 1, 3), (6, 1, 4), (6, 1, 5), (6, 2, 1), (6, 2, 3), (6, 2, 4), (6, 2, 5), (6, 3, 1), (6, 3, 2), (6, 3, 4), (6, 3, 5), (6, 4, 1), (6, 4, 2), (6, 4, 3), (6, 4, 5), (6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4)]

这也不是很好，因为它只会产生双倍或三倍的结果，就我所知，一切都只是唯一的组合。在

------------更新----------- 好的——我取了这个，通过从数组中剥离每个值并求和（可能以最低效的方式）对其进行了扩展。它可以工作，如果在中断之前生成了多个集合，它们都会打印出来。我现在想做的是求和。所以：

^{pr2}$

下面是输出示例：

How many tervigons? ::>3
Let's calculate some termagants based on 3 tervigons...
You'll get a minimum of 9 termagants per turn.
You'll get a maximum of 54 termagants per turn.
minimums: 5 turns [45] :: 6 turns [54] :: 7 turns [63]
averages: 5 turns [157] :: 6 turns [189] :: 7 turns [220]
maximums: 5 turns [270] :: 6 turns [324] :: 7 turns [378]
Total: 9
Total: 8

所以在这个例子中，我希望它返回17（即9+8）。在

Answer 1

Python附带了一个great standard library（您可能已经发现了，因为您正在使用itertools），在这个模块中您还可以找到一个random模块。在

你可以用random.randint来模拟掷骰子。有多种方法可以解决这个问题。第一个代码示例有点有限，第二个更通用。在

import random

# '_' (underscore) is used for values that are generated, but that you do not
# care about - here we only want to repeat seven times and do not care about
# the actual loop count 
for _ in range(7): 
    # generate three random numbers between [1 and 6] 
    # and store the values in a, b, c respectively (tuple unpacking)
    a, b, c = (random.randint(1, 6) for _ in range(3))

    # if one of the conditions holds, break out of the loop early
    if a == b or a == c or b == c or a == b == c:
        break

正如@Paulo所指出的，您可以使用另一种更简洁的方法来检查列表（或元组）的n元素是否都不同，即将所有元素放入set：

^{pr2}$

回答更新后的问题，只需使用^{}：

    total = sum(dice)

Answer 2

这是输入排列的正确输出，你要的是笛卡尔积。见product。在

这将打印出您期望的输出。在

from itertools import product

d = [ x for x in range(1, 7) ]
for r in product(d, d, d):
    print r