Question

我尝试下面的代码，有没有一种有效的方法来做到这一点？

c = []
l = [['A1','A2'], ['B1','B2'],  ['C1','C2'] ]

for i in range(0, len(l) - 1):
    for j in range(i+1, len(l)): 
        c.append(sorted([l[i][0],l[i][1],l[j][0]]))
        c.append(sorted([l[i][0],l[i][1],l[j][1]]))
        c.append(sorted([l[i][0],l[j][0],l[j][1]]))
        c.append(sorted([l[i][1],l[j][0],l[j][1]]))

print(c)

Out put：

[['A1', 'A2', 'B1'], ['A1', 'A2', 'B2'], ['A1', 'B1', 'B2'],
['A2', 'B1', 'B2'], ['A1', 'A2', 'C1'], ['A1', 'A2', 'C2'], 
['A1', 'C1', 'C2'], ['A2', 'C1', 'C2'], ['B1', 'B2', 'C1'], 
['B1', 'B2', 'C2'], ['B1', 'C1', 'C2'], ['B2', 'C1', 'C2']

Answer 1

试试这个：

# group every 2 lists in list l
ll = list(itertools.combinations(l, 2))

# generate all combinations of 3 elements out from each 2 lists
c = [list(itertools.combinations(a + b, 3)) for (a, b) in ll]

# concate all elements
c = sum(c, [])

Answer 2

或在一行

from itertools import product

c = [[k] + i for i, j in product(l, l) if j!=i for k in j]

Python与列表和其他列表中的项目的组合

2 个答案: