Python：随机从多的名单，没有repeats


Python: Random from multiple lists, no repeats

本问题已经有最佳答案，请猛点这里访问。

假设我有4个列表，每个列表有10个不同的选项，我想从四个列表中选择一个项目，用户指定的次数。但没有重复的输出。这有可能吗？我似乎写不出任何不会重复输出的东西。

相关讨论

假设我们有这些4列表：

1
2
3
>>> lists = [range(10*i, 10*(i+1)) for i in range(4)]
>>> lists
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [20, 21, 22, 23, 24, 25, 26, 27, 28, 29], [30, 31, 32, 33, 34, 35, 36, 37, 38, 39]]

首先，洗牌：

1
2
3
4
>>> for lst in lists: random.shuffle(lst)

>>> lists
[[6, 8, 2, 1, 3, 5, 9, 0, 7, 4], [17, 12, 16, 10, 14, 15, 18, 11, 13, 19], [20, 28, 23, 21, 27, 25, 24, 29, 26, 22], [35, 32, 38, 31, 39, 34, 30, 33, 36, 37]]

然后用zip输出值：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
>>> for items in zip(*lists):
    print(items)


(6, 17, 20, 35)
(8, 12, 28, 32)
(2, 16, 23, 38)
(1, 10, 21, 31)
(3, 14, 27, 39)
(5, 15, 25, 34)
(9, 18, 24, 30)
(0, 11, 29, 33)
(7, 13, 26, 36)
(4, 19, 22, 37)

如果您只需要指定的数量，只需使用islice：

1
2
3
4
5
6
7
8
9
10
>>> from itertools import islice
>>> for items in islice(zip(*lists),5):
    print(items)


(6, 17, 20, 35)
(8, 12, 28, 32)
(2, 16, 23, 38)
(1, 10, 21, 31)
(3, 14, 27, 39)
1
2
3
population = [1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1]
population = set(population)
samples = random.sample(population, 3)

这样的方法应该有效：

1
2
3
4
5
6
7
8
9
10
a,b,c,d = initialize_lists()
output_picked_before = set()
n = number_of_experiments()

for i in range(n):
    t = pick_random_combination_from_list(a,b,c,d)
    while t in output_picked_before:
        t = pick_random_combination_from_list(a,b,c,d)
    print t
    output_picked_before.add(t)