如何获得 2 个列表的所有组合？

Question

我如何在 Python 中执行此操作？输入：num = [1, 2, 3]alpha = ['a', 'b', 'c']输出：[1, a] [1, a, b] [1, a, b, c] [1, b, c] [1, c][1, 2, a] [1, 2, a, b] ... [1, 2, c]......[3, a] [3, a, b] ... [3, c]约束：The output expect all possible consecutive combinationse.g. ['a'] ['a','b'] are desired while ['a','c'] is not.我尝试了 4 个 for 循环/4 个 while 循环的嵌套。使用这种深度的循环是否常见？

Answer 1

以下仅使用两个嵌套的 for 循环

from itertools import product as prod

def consecutive_combos(a):

    " Generates consecutive ombinations of items in list "

    return [a[i:j] for i in range(len(a)) for j in range(i+1, len(a)+1)]

num = [1, 2, 3]

alpha = ['a', 'b', 'c']

# Generates product of sequences

result = [x + y for x, y in prod(consecutive_combos(num), consecutive_combos(alpha))]

print(result)

输出

[[1, 'a'], [1, 'a', 'b'], [1, 'a', 'b', 'c'], [1, 'b'], [1, 'b', 'c'], [1, 'c'], [1, 2, 'a'], [1, 2, 'a', 'b'], [1, 2, 'a', 'b', 'c'], [1, 2, 'b'], [1, 2, 'b', 'c'], [1, 2, 'c'], [1, 2, 3, 'a'], [1, 2, 3, 'a', 'b'], [1, 2, 3, 'a', 'b', 'c'], [1, 2, 3, 'b'], [1, 2, 3, 'b', 'c'], [1, 2, 3, 'c'], [2, 'a'], [2, 'a', 'b'], [2, 'a', 'b', 'c'], [2, 'b'], [2, 'b', 'c'], [2, 'c'], [2, 3, 'a'], [2, 3, 'a', 'b'], [2, 3, 'a', 'b', 'c'], [2, 3, 'b'], [2, 3, 'b', 'c'], [2, 3, 'c'], [3, 'a'], [3, 'a', 'b'], [3, 'a', 'b', 'c'], [3, 'b'], [3, 'b', 'c'], [3, 'c']]

Answer 2

您可以使用模块itertoolsanscombinations

import itertools as it

l1 = []

# get all combinations of num

for i in range(1, len(num)+1):

    l1.extend([* it.combinations(num, i)])

# get all combinations of alpha

for j in range(1, len(alpha)+1):

    l2.extend([* it.combinations(alpha, j)])

# list comprehension to combine elements from the two lists

comb = [e1+e2 for e1 in l1 for e2 in l2]       

通过这样做，您将获得一个元组列表。

更新：

为了考虑约束：

输出期望所有可能的连续组合

例如 [a] [a,b] 是需要的，而 [a,c] 不是。

num = [1, 2, 3]

alpha = ['a', 'b', 'c']

l1 = [num[i:j+1] for i in range(len(num)) for j in range(i, len(num))]

l2 = [alpha[i:j+1] for i in range(len(alpha)) for j in range(i, len(alpha))]

result = [e1+e2 for e1 in l1 for e2 in l2]

print(*result, sep="\n")

[1, 'a']

[1, 'a', 'b']

[1, 'a', 'b', 'c']

[1, 'b']

[1, 'b', 'c']

[1, 'c']

[1, 2, 'a']

[1, 2, 'a', 'b']

[1, 2, 'a', 'b', 'c']

[1, 2, 'b']

[1, 2, 'b', 'c']

[1, 2, 'c']

[1, 2, 3, 'a']

[1, 2, 3, 'a', 'b']

[1, 2, 3, 'a', 'b', 'c']

[1, 2, 3, 'b']

[1, 2, 3, 'b', 'c']

[1, 2, 3, 'c']

[2, 'a']

[2, 'a', 'b']

[2, 'a', 'b', 'c']

[2, 'b']

[2, 'b', 'c']

[2, 'c']

[2, 3, 'a']

[2, 3, 'a', 'b']

[2, 3, 'a', 'b', 'c']

[2, 3, 'b']

[2, 3, 'b', 'c']

[2, 3, 'c']

[3, 'a']

[3, 'a', 'b']

[3, 'a', 'b', 'c']

[3, 'b']

[3, 'b', 'c']

[3, 'c']

Answer 3

我希望它会有所帮助

from itertools import combinations 

num = [1, 2, 3]

alpha = ["a", "b", "c"]

big_list=num+alpha

comb += list(combinations(big_list,2) )

comb += list(combinations(big_list,3) )

for i in list(comb): 

   print (i)

输出：

(1, 2) (1, 3) (1, 'a') (1, 'b') (1, 'c') (2, 3) (2, 'a') (2, 'b') (2, 'c') (3, 'a') (3, 'b') (3, 'c') ('a', 'b') ('a', 'c') ('b' , 'c') (1, 2, 3) (1, 2, 'a') (1, 2, 'b') (1, 2, 'c') (1, 3, 'a') (1 , 3, 'b') (1, 3, 'c') (1, 'a', 'b') (1, 'a', 'c') (1, 'b', 'c') ( 2, 3, 'a') (2, 3, 'b') (2, 3, 'c') (2, 'a', 'b') (2, 'a', 'c') (2 , 'b', 'c') (3, 'a', 'b') (3, 'a', 'c') (3, 'b', 'c') ('a', 'b' , 'c') [0.50秒内完成]