如何从不同列表中识别3个图层组合 [英] How to identify 3 layer combinations from different lists

问题描述

我有两个不同的图像数据集列表(列表1和列表2).考虑到两个列表的集成，我想确定可能的三种功能组合，但不能单独考虑.list1是主要数据集，list2是辅助数据.我已经熟悉 itertools.combinations ，但是我不知道如何解决当前问题.谁能在python中提出解决方案?

I have two different lists (list1 and list2) of image datasets. I want to determine possible three feature combinations considering the integration of both lists, but not individually. list1 is the main dataset and list2 is ancillary data. I am already familiar with itertools.combinations but I don't know how to solve the current problem. Can anyone suggest a solution in python?

list1= ["a","b","c","d","e"]

list2= ["a1", "a2","a3","a4",
        "a5","a6","a7","a8",
        "a9","a10","a11","a12"]

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五个可能组合的示例:

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example for five possible combinations:

combinations=[('a', 'a1', 'a2'), ('a', 'b', 'a2'), ('b', 'a1', 'a2'), ('b', 'c', 'a3'), ('a', 'a2', 'a3')]

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我测试了建议的解决方案的两个小列表，这会导致某些组合中成员的重复.

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I tested the proposed solution for two small lists and this produce repetition of members in some combinations.

list1= ['a','b','c','d','e']
list2= ['a1', 'a2']

结果:

{('d', 'c', 'a2'), ('c', 'a2', 'a1'), ('e', 'b', 'a1'), ('c', 'd', 'a1'), ('e', 'a2', 'a1'), ('a', 'a', 'a2'), ('b', 'a2', 'a1'), ('d', 'd', 'a1'), ('d', 'a', 'a2'), ('a', 'b', 'a2'), ('a', 'e', 'a2'), ('e', 'a1', 'a1'), ('d', 'e', 'a2'), ('c', 'a', 'a2'), ('c', 'c', 'a2'), ('c', 'e', 'a2'), ('b', 'd', 'a1'), ('a', 'c', 'a1'), ('e', 'd', 'a1'), ('d', 'b', 'a2'), ('e', 'c', 'a2'), ('d', 'c', 'a1'), ('b', 'c', 'a2'), ('b', 'e', 'a2'), ('c', 'b', 'a2'), ('a', 'a', 'a1'), ('a', 'a1', 'a2'), ('d', 'a', 'a1'), ('a', 'e', 'a1'), ('a', 'b', 'a1'), ('b', 'a', 'a2'), ('d', 'a1', 'a2'), ('d', 'e', 'a1'), ('e', 'e', 'a2'), ('a', 'a2', 'a2'), ('c', 'a', 'a1'), ('c', 'c', 'a1'), ('a', 'd', 'a2'), ('d', 'a2', 'a2'), ('c', 'a1', 'a2'), ('c', 'e', 'a1'), ('b', 'a1', 'a1'), ('d', 'b', 'a1'), ('e', 'a', 'a2'), ('e', 'c', 'a1'), ('b', 'b', 'a2'), ('b', 'a1', 'a2'), ('b', 'c', 'a1'), ('e', 'b', 'a2'), ('c', 'a2', 'a2'), ('c', 'd', 'a2'), ('b', 'e', 'a1'), ('c', 'b', 'a1'), ('e', 'a2', 'a2'), ('a', 'a1', 'a1'), ('d', 'd', 'a2'), ('b', 'a2', 'a2'), ('b', 'a', 'a1'), ('e', 'a1', 'a2'), ('d', 'a1', 'a1'), ('e', 'e', 'a1'), ('a', 'a2', 'a1'), ('a', 'd', 'a1'), ('e', 'd', 'a2'), ('d', 'a2', 'a1'), ('c', 'a1', 'a1'), ('b', 'd', 'a2'), ('a', 'c', 'a2'), ('e', 'a', 'a1'), ('b', 'b', 'a1')}

重复的例子:('c'，'a1'，'a1')

推荐答案

收集所有所需信息后，我建议的解决方案是使用 product :

After gathering all the information required, my proposed solution is to use a product instead:

from itertools import product

def apply_op(list1, list2):
   part_1 = product(list1, list1, list2)
   part_2 = product(list1, list2, list2)
   return set(list(part_1)).union(set(list(part_2)))

print(apply_op(A, B))

以上解决方案不起作用，因为结果在元组中包含重复的成员.

above solution doesn't work, since the result contains duplicate members in tuples.

def apply_op(list1, list2):
    ret = []
    for i in range(len(list1)):
        for j in range(i + 1, len(list1)):
            for k in range(len(list2)):
                ret.append((list1[i], list1[j], list2[k]))
    for i in range(len(list1)):
        for j in range(len(list2)):
            for k in range(j + 1, len(list2)):
                ret.append((list1[i], list2[j], list2[k]))
    return ret

print(apply_op(A, B))

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