帶有可選列表的笛卡爾積

Question

我正在 python 中創建一個程序，它允許我根據給定的資產生成 NFT 藝術。 顯然，可以生成的藝術數量根據資產（圖層和圖層圖像）而變化，這正是問題所在，我如何計算可能的組合並計算可選圖層？

更清楚一點：

例如我有 4 層：

l1 = ["A","B"]
l2 = ["C"]
l3 = ["D","E"] #optional
l4 = ["F","G"] #optional

其中 l3 和 l4 是可選的。 所以我期望的組合是：

 1.  ["A","C"]
 2.  ["B","C"]
 3.  ["A","C","D"]
 4.  ["A","C","E"]
 5.  ["B","C","D"]
 6.  ["B","C","E"]
 7.  ["A","C","F"]
 8.  ["A","C","G"]
 9.  ["B","C","F"]
 10. ["B","C","G"]
 11. ["A","C","D","F"]
 12. ["A","C","D","G"]
 13. ["A","C","E","F"]
 14. ["A","C","E","G"]
 15. ["B","C","D","F"]
 16. ["B","C","D","G"]
 17. ["B","C","E","F"]
 18. ["B","C","E","G"]

我怎樣才能做到這一點？ 我嘗試使用itertools.product但顯然它考慮了所有列表

Answer 1

我假設可選層的順序很重要，因此您可以迭代創建可選層的所有組合，然后在layers上使用itertools.product + optional_layers來生成列表。

import itertools
from pprint import pprint

l1 = ["A","B"]
l2 = ["C"]
l3 = ["D","E"] #optional
l4 = ["F","G"] #optional

layers = [l1, l2]
optional_layers = [l3, l4]

results = []
results += itertools.product(*layers)
for i in range(len(optional_layers) + 1):
    comb = itertools.combinations(optional_layers, r=i)
    for c in comb:
        results += itertools.product(*layers, *c)

pprint(results)

Output

[('A', 'C'),
 ('B', 'C'),
 ('A', 'C'),
 ('B', 'C'),
 ('A', 'C', 'D'),
 ('A', 'C', 'E'),
 ('B', 'C', 'D'),
 ('B', 'C', 'E'),
 ('A', 'C', 'F'),
 ('A', 'C', 'G'),
 ('B', 'C', 'F'),
 ('B', 'C', 'G'),
 ('A', 'C', 'D', 'F'),
 ('A', 'C', 'D', 'G'),
 ('A', 'C', 'E', 'F'),
 ('A', 'C', 'E', 'G'),
 ('B', 'C', 'D', 'F'),
 ('B', 'C', 'D', 'G'),
 ('B', 'C', 'E', 'F'),
 ('B', 'C', 'E', 'G')]

Answer 2

一種方法是使用itertools 文檔中的powerset配方。 將“必需列表”的產品與“可選列表集”的每個子集鏈接在一起，生成一個生成每種可能性一次的生成器：

def powerset(iterable):
    """powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"""
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s) + 1))

def product_with_optional(required_sequences, optional_sequences):
    return chain.from_iterable(product(*required_sequences, *optionals)
                               for optionals in powerset(optional_sequences))

optional_combinations = product_with_optional(required_sequences=[l1, l2],
                                              optional_sequences=[l3, l4])

這使：

1 ('A', 'C')
2 ('B', 'C')
3 ('A', 'C', 'D')
4 ('A', 'C', 'E')
5 ('B', 'C', 'D')
6 ('B', 'C', 'E')
7 ('A', 'C', 'F')
8 ('A', 'C', 'G')
9 ('B', 'C', 'F')
10 ('B', 'C', 'G')
11 ('A', 'C', 'D', 'F')
12 ('A', 'C', 'D', 'G')
13 ('A', 'C', 'E', 'F')
14 ('A', 'C', 'E', 'G')
15 ('B', 'C', 'D', 'F')
16 ('B', 'C', 'D', 'G')
17 ('B', 'C', 'E', 'F')
18 ('B', 'C', 'E', 'G')