生成串列，允許重復并符合條件

從選擇 [1,2,3] 中，我想輸出所有可能的組合，允許選擇重復，在 5 個元素的串列中。

在每個串列中，必須至少有 1 個，至少有 2 個，至少有 3 個。

一個笨拙的方法如下。它首先使用 [1,2,3] 中的任一個生成一個 5 的串列。檢查所有生成的串列，以確保至少具有 [1,2,3] 中的每一個。合格的被列入一個大名單。然后洗掉大串列中的重復項（回圈多次以確保覆寫良好）：

import random
import itertools

choices = [1,2,3]

big_list = []

for a in range(10000):
    new_list = [random.choice(choices) for i in range(5)]

    if new_list.count(1) >= 1 and new_list.count(2) >= 1 and new_list.count(3) >= 1:
        big_list.append(new_list)

big_list.sort()

final_list = list(big_list for big_list, _ in itertools.groupby(big_list))
# this line to remove the duplicates in the list of lists

print (final_list)

考慮到順序問題，即 [1,1,1,2,3] 和 [2,3,1,1,1] 是兩個不同的串列。

這樣做的更聰明和更全面的方法是什么？謝謝你。

uj5u.com熱心網友回復：

也許您可以使用,以及：itertools.combinations_with_replacementitertools.permutationscollections.Counter

>>> from collections import Counter
>>> from itertools import combinations_with_replacement, permutations
>>> 
>>> def is_valid_combination(comb: tuple) -> bool:
...     digit_counts = Counter(comb)
...     return digit_counts[1] >= 1 and \
...            digit_counts[2] >= 1 and \
...            digit_counts[3] >= 1
... 
>>> choices = [1, 2, 3]
>>> valid_combinations = [
...     c for c in combinations_with_replacement(choices, r=5)
...     if is_valid_combination(c)
... ]
>>> 
>>> valid_combinations
[(1, 1, 1, 2, 3), (1, 1, 2, 2, 3), (1, 1, 2, 3, 3), (1, 2, 2, 2, 3), (1, 2, 2, 3, 3), (1, 2, 3, 3, 3)]
>>>
>>> all_permutations_of_valid_combinations = {
...     p
...     for c in valid_combinations for p in permutations(c)
... }
>>>
>>> all_permutations_of_valid_combinations
{(2, 1, 3, 1, 2), (2, 1, 3, 2, 1), (3, 3, 2, 1, 3), (1, 2, 3, 2, 3), (1, 2, 1, 3, 1), (3, 1, 2, 3, 2), (3, 3, 3, 2, 1), (3, 2, 2, 1, 1), (1, 2, 2, 3, 1), (1, 3, 2, 2, 3), (1, 3, 2, 3, 2), (1, 2, 1, 1, 3), (3, 1, 3, 3, 2), (3, 1, 1, 2, 3), (2, 1, 3, 2, 3), (1, 2, 2, 1, 3), (1, 2, 1, 3, 3), (2, 3, 3, 1, 2), (2, 3, 3, 2, 1), (3, 3, 1, 2, 1), (3, 2, 3, 2, 1), (1, 2, 2, 3, 3), (3, 2, 1, 1, 1), (2, 2, 1, 3, 1), (2, 3, 1, 1, 1), (1, 3, 1, 2, 3), (3, 3, 1, 1, 2), (3, 2, 3, 1, 2), (2, 1, 2, 3, 1), (2, 2, 1, 1, 3), (3, 2, 1, 3, 1), (2, 3, 1, 3, 1), (1, 1, 3, 2, 1), (2, 3, 2, 1, 2), (2, 3, 2, 2, 1), (2, 1, 2, 1, 3), (3, 2, 1, 1, 3), (2, 2, 1, 3, 3), (2, 3, 1, 1, 3), (2, 3, 1, 2, 2), (3, 2, 3, 3, 1), (1, 1, 3, 1, 2), (2, 1, 2, 3, 3), (3, 3, 2, 2, 1), (3, 1, 2, 1, 2), (3, 2, 1, 3, 3), (3, 1, 2, 2, 1), (2, 3, 1, 3, 3), (1, 1, 3, 2, 3), (3, 3, 3, 1, 2), (1, 2, 3, 1, 1), (1, 1, 3, 3, 2), (3, 1, 3, 1, 2), (2, 3, 2, 3, 1), (1, 3, 2, 1, 1), (2, 1, 3, 3, 1), (3, 2, 2, 3, 1), (3, 1, 2, 2, 3), (1, 3, 2, 2, 2), (1, 2, 3, 1, 3), (1, 3, 2, 3, 1), (3, 2, 2, 1, 3), (2, 2, 3, 2, 1), (3, 1, 1, 2, 2), (1, 1, 2, 2, 3), (2, 1, 3, 2, 2), (1, 3, 3, 2, 2), (3, 3, 1, 3, 2), (2, 1, 1, 3, 1), (1, 3, 2, 1, 3), (2, 1, 3, 3, 3), (3, 1, 3, 2, 2), (2, 2, 3, 1, 2), (1, 1, 2, 3, 1), (3, 2, 1, 2, 2), (1, 2, 2, 3, 2), (3, 3, 1, 2, 3), (1, 3, 2, 3, 3), (1, 2, 1, 2, 3), (3, 2, 3, 1, 1), (1, 3, 1, 2, 2), (1, 2, 2, 2, 3), (2, 1, 1, 3, 3), (3, 1, 1, 3, 2), (1, 1, 2, 3, 3), (1, 3, 3, 3, 2), (2, 3, 2, 1, 1), (2, 2, 1, 2, 3), (2, 2, 1, 3, 2), (1, 2, 3, 3, 1), (3, 2, 3, 1, 3), (2, 3, 1, 2, 1), (2, 1, 3, 1, 1), (3, 3, 2, 1, 2), (1, 2, 3, 2, 2), (1, 3, 1, 3, 2), (3, 1, 2, 3, 1), (2, 2, 2, 3, 1), (2, 1, 2, 2, 3), (1, 2, 3, 3, 3), (2, 3, 1, 2, 3), (2, 1, 3, 1, 3), (3, 2, 2, 2, 1), (1, 2, 1, 3, 2), (2, 3, 3, 1, 1), (3, 1, 2, 3, 3), (3, 2, 2, 1, 2), (3, 1, 1, 2, 1), (1, 3, 3, 2, 1), (2, 3, 3, 3, 1), (2, 1, 1, 1, 3), (1, 3, 2, 1, 2), (2, 1, 3, 3, 2), (1, 1, 1, 2, 3), (3, 1, 3, 2, 1), (1, 1, 1, 3, 2), (2, 2, 3, 1, 1), (3, 1, 1, 1, 2), (1, 1, 2, 1, 3), (1, 3, 3, 1, 2), (3, 2, 1, 2, 1), (2, 3, 3, 1, 3), (3, 3, 1, 2, 2), (2, 2, 3, 3, 1), (1, 3, 1, 2, 1), (1, 3, 3, 2, 3), (3, 2, 1, 1, 2), (2, 1, 1, 3, 2), (2, 3, 1, 1, 2), (3, 1, 3, 2, 3), (2, 2, 3, 1, 3), (1, 3, 1, 1, 2), (1, 1, 2, 3, 2), (2, 1, 2, 3, 2), (3, 2, 1, 2, 3), (3, 1, 2, 1, 1), (3, 2, 1, 3, 2), (2, 1, 1, 2, 3), (2, 3, 1, 3, 2), (1, 1, 3, 2, 2), (2, 3, 2, 1, 3), (3, 3, 2, 3, 1), (3, 3, 2, 1, 1), (1, 2, 3, 2, 1), (3, 1, 2, 1, 3), (2, 2, 2, 1, 3), (3, 1, 2, 2, 2), (1, 3, 2, 2, 1), (1, 2, 3, 1, 2), (1, 2, 3, 3, 2)}

uj5u.com熱心網友回復：

除了itertools.combinations它本身，您還可以使用一些遞回邏輯：

def combinations(choices, n = 5):
    if n == 1:
        return [[x,] for x in choices]
    else:
        return [v   [x,] for v in combinations(choices, n = n -1) for x in choices]

僅選擇包含至少 1、2 和 3 之一的組合：

[x for x in combinations(choices, n = 5) if all(c in x for c in choices)]

標籤：Python 列表 随机的 组合

上一篇：STM32+W5500網路通信

下一篇：3V3TTL和5VTTL沒做轉換, 不能通訊