给定 2 个列表，从两个列表中找到一个随机固定大小的子集，使得每个列表中至少有一个值（最好是统一选择的）

Question

Let's say I have 2 lists, list 1 and list 2 such that( for example)

l1 = ['w1', 'w2', 'w3', 'w4', 'w5']
l2 = ['w6', 'w7', 'w8']

Is there a short and sweet way to get a new list (of a given fixed size, such as 4) containing words from both lists such that there is at least one word from each list? (No repetition)

Possible results for size 4

['w6', 'w2', 'w4', 'w8']
['w2', 'w8', 'w7', 'w4']
['w1', 'w2', 'w6', 'w4']
['w2', 'w3', 'w1', 'w7']

Answer 1

You could use random.sample() for that:

import random

l1 = ['w1','w2','w3','w4','w5']
l2 = ['w6','w7','w8']

result = [random.sample(l1,2) + random.sample(l2,2) for i in range(4)]
print(result)

Possible result:

[['w5', 'w1', 'w8', 'w7'], ['w3', 'w4', 'w7', 'w6'], ['w3', 'w5', 'w6', 'w8'], ['w5', 'w2', 'w7', 'w6']]

Answer 2

You can generate all of them:

from itertools import combinations

l1 = ['w1','w2','w3','w4','w5']
l2 = ['w6','w7','w8']

results = []
for parts in ( list(p) + [other] for p in combinations(l1,3) for other in l2):
    results.append(parts)

print(results, sep="\n")

Output:

[['w1', 'w2', 'w3', 'w6'], ['w1', 'w2', 'w3', 'w7'], ['w1', 'w2', 'w3', 'w8'],
 ['w1', 'w2', 'w4', 'w6'], ['w1', 'w2', 'w4', 'w7'], ['w1', 'w2', 'w4', 'w8'], 
 ['w1', 'w2', 'w5', 'w6'], ['w1', 'w2', 'w5', 'w7'], ['w1', 'w2', 'w5', 'w8'],
 ['w1', 'w3', 'w4', 'w6'], ['w1', 'w3', 'w4', 'w7'], ['w1', 'w3', 'w4', 'w8'],
 ['w1', 'w3', 'w5', 'w6'], ['w1', 'w3', 'w5', 'w7'], ['w1', 'w3', 'w5', 'w8'],
 ['w1', 'w4', 'w5', 'w6'], ['w1', 'w4', 'w5', 'w7'], ['w1', 'w4', 'w5', 'w8'],
 ['w2', 'w3', 'w4', 'w6'], ['w2', 'w3', 'w4', 'w7'], ['w2', 'w3', 'w4', 'w8'],
 ['w2', 'w3', 'w5', 'w6'], ['w2', 'w3', 'w5', 'w7'], ['w2', 'w3', 'w5', 'w8'],
 ['w2', 'w4', 'w5', 'w6'], ['w2', 'w4', 'w5', 'w7'], ['w2', 'w4', 'w5', 'w8'],
 ['w3', 'w4', 'w5', 'w6'], ['w3', 'w4', 'w5', 'w7'], ['w3', 'w4', 'w5', 'w8']]

- itertools.combinations of l1 generates all 3-long combinations of l1 and adds one element of l2 to it.

Answer 3

You can combine the lists and use a generator function:

l1 = ['w1', 'w2', 'w3', 'w4', 'w5']
l2 = ['w6', 'w7', 'w8']
def combos(d, c = []):
  if len(c) == 4:
    yield c
  else:
    for i in d:
       s1, s2 = sum(i in c for i in l1), sum(i in c for i in l2)
       if not (s1 and s2) and len(c) == 3:
          if i not in c and ((not s1 and i in l1) or (not s2 and i in l2)):
             yield from combos(d, c+[i])
       elif i not in c:
           yield from combos(d, c+[i])

print(list(combos(l1+l2)))

Output:

[['w1', 'w2', 'w3', 'w6'], 
 ['w1', 'w2', 'w3', 'w7'], 
 ['w1', 'w2', 'w3', 'w8'], 
 ['w1', 'w2', 'w4', 'w6'], 
 ['w1', 'w2', 'w4', 'w7'], 
 ['w1', 'w2', 'w4', 'w8']
 ....
 ['w6', 'w1', 'w7', 'w3'], 
 ['w6', 'w1', 'w7', 'w4'], 
 ['w6', 'w1', 'w7', 'w5'], 
 ['w6', 'w1', 'w7', 'w8'], 
 ['w6', 'w1', 'w8', 'w2']
 ....
 ]