问题：我需要掷3个骰子。如果两个（或三个）骰子返回相同的数字，则停止。如果3个骰子都是唯一的（例如2,4和6），则再次滚动。执行此操作，直到双打/三重滚动，或7次，以先到者为准。
注意：我是一个python newb。
这是我到目前为止所做的，但这一切实际上产生了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
这会产生这种类型的输出：
[(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，因为它只产生NO双打或三倍 - 据我所知，一切都只是唯一的组合。
---------- -----------更新 好的 - 我拿了这个并通过从数组中剥离每个值并将它们相加（可能以最低效的方式）来扩展它。它有效，如果在休息之前有多个集合，它们都会打印出来。我现在要做的就是总结。所以：
def gen_random_termagants():
for _ in range(7):
# instead of three separate variables, we use a list here
# the nice thing is, that you can freely vary the number of
# 'parallel' dice rolls this way
dice = [random.randint(1, 6) for _ in range(3)]
# this is more general and will break as soon as there are
# duplicate (die) values in the list (meaning, break, if not all elements
# are different)
first_die = dice[0]
second_die = dice[1]
third_die = dice[2]
total_term = first_die + second_die + third_die
print "Total: %s" % total_term
if len(dice) > len(set(dice)):
break
这是输出样本：
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）。
答案 0 :(得分：7)
Python附带一个great standard library（因为您使用 itertools 时可能已经发现过），您还可以在其中找到random模块。
您可以使用random.randint来模拟掷骰子。有多种方法可以解决这个问题。第一个代码示例有些限制，第二个代码示例更为通用。
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指出的那样，您可以使用另一种更简洁的方法来检查列表（或元组）的n元素是否全部不同，即您放置所有元素set中的元素：
for _ in range(7):
# instead of three separate variables, we use a list here
# the nice thing is, that you can freely vary the number of
# 'parallel' dice rolls this way
dice = [random.randint(1, 6) for _ in range(3)]
# this is more general and will break as soon as there are
# duplicate (die) values in the list (meaning, break, if not all elements
# are different)
if len(dice) > len(set(dice)):
break
回答您的更新问题，只需使用sum：
total = sum(dice)
答案 1 :(得分：1)
这是输入排列的正确输出，您要求的是笛卡尔积。请参阅product。
E.G。这将打印您期望的输出。
from itertools import product
d = [ x for x in range(1, 7) ]
for r in product(d, d, d):
print r