如何从第一个列表中选择一个项目，从第二个列表中最多找出n个项目的所有组合。

Question

我有两个列表，我想返回一个新的列表，其中每个列表都是从第一个列表中选择一个项目，而最多是从第二个列表中选择一个2项:例如：

vaccinate_list: [('vaccinate', (2, 3)), ('vaccinate', (8, 1))]
quarantine_list: [('quarantine', (6, 3)), ('quarantine', (3, 4)), ('quarantine', (0, 1))]

将返回(没有新行)：

allCombinations = [
[('vaccinate', (2, 3))], 
[('vaccinate', (8, 1))], 
[('vaccinate', (2, 3)), ('quarantine', (6, 3))], 
[('vaccinate', (2, 3)), ('quarantine', (3, 4))], 
[('vaccinate', (2, 3)), ('quarantine', (0, 1))], 
[('vaccinate', (8, 1)), ('quarantine', (6, 3))], 
[('vaccinate', (8, 1)), ('quarantine', (3, 4))], 
[('vaccinate', (8, 1)), ('quarantine', (0, 1))], 
[('vaccinate', (2, 3)), ('quarantine', (6, 3)), ('quarantine', (3, 4))], 
[('vaccinate', (2, 3)), ('quarantine', (6, 3)), ('quarantine', (0, 1))], 
[('vaccinate', (2, 3)), ('quarantine', (3, 4)), ('quarantine', (0, 1))], 
[('vaccinate', (8, 1)), ('quarantine', (6, 3)), ('quarantine', (3, 4))], 
[('vaccinate', (8, 1)), ('quarantine', (6, 3)), ('quarantine', (0, 1))], 
[('vaccinate', (8, 1)), ('quarantine', (3, 4)), ('quarantine', (0, 1))]
]

我尝试了工具间的组合，然后挣扎着，请帮助。

Answer 1

最好的方法是使用itertools.product。见下文(评论中的解释)：

allCombinations=[]

allCombinations.extend(vaccinate_list) #add items of vaccinate_list only

list_one=[i for i in itertools.product(vaccinate_list, quarantine_list)] #get all combinations with one item from each list

allCombinations.extend(list_one) #add them to allcombinations

list_two_temp=[i for i in itertools.combinations(quarantine_list, 2)] #get combinations of 2, in quarantine_list

list_two=[i for i in itertools.product(vaccinate_list, list_two_temp)] #get combinations with 1 item from vaccinate_list and 2 from quarantine_list

list_two=[(i[0],i[1][0], i[1][1]) for i in list_two] #flatten the items from previous list

allCombinations.extend(list_two) #add them to allcombinations

allCombinations=[list(i) for i in allCombinations] #change all items to list

print(allCombinations)

输出：

[['vaccinate', (2, 3)], ['vaccinate', (8, 1)], [('vaccinate', (2, 3)), ('quarantine', (6, 3))], [('vaccinate', (2, 3)), ('quarantine', (3, 4))], [('vaccinate', (2, 3)), ('quarantine', (0, 1))], [('vaccinate', (8, 1)), ('quarantine', (6, 3))], [('vaccinate', (8, 1)), ('quarantine', (3, 4))], [('vaccinate', (8, 1)), ('quarantine', (0, 1))], [('vaccinate', (2, 3)), ('quarantine', (6, 3)), ('quarantine', (3, 4))], [('vaccinate', (2, 3)), ('quarantine', (6, 3)), ('quarantine', (0, 1))], [('vaccinate', (2, 3)), ('quarantine', (3, 4)), ('quarantine', (0, 1))], [('vaccinate', (8, 1)), ('quarantine', (6, 3)), ('quarantine', (3, 4))], [('vaccinate', (8, 1)), ('quarantine', (6, 3)), ('quarantine', (0, 1))], [('vaccinate', (8, 1)), ('quarantine', (3, 4)), ('quarantine', (0, 1))]]

Answer 2

for i in vaccinate_list:
    all_combinations.append(i)
for i in vaccinate_list:
    for j in quarantine_list:
         all_combinations.append((i,j))
for i in vaccinate_list:
     for j in quarantine_list:
          for k in quarantine_list:
               all_combinations.append((i, j, k))

可能很复杂，也可能是一种愚蠢的方法，但它有效。