从可能包含子列表的元素列表中生成k个随机列表

Question

I have a list l in the following form. I need to randomly generate k (six in this example) number of lists from this list so that only one element is selected from the sublists at a time.

l = [1,2,3,[11,22,33,44], 4,5,6, [22,33,44], 5, [99,88]]

Result: 

1,2,3, 22, 4,5,6, 22 ,5, 88
1,2,3, 33, 4,5,6, 44 ,5, 88
1,2,3, 44, 4,5,6, 22 ,5, 99
1,2,3, 22, 4,5,6, 33 ,5, 99
1,2,3, 33, 4,5,6, 33 ,5, 99
1,2,3, 33, 4,5,6, 44 ,5, 88

I can write a for loop and pick a random element whenever I encountered a list. But i am looking for more elegant pythonic way to do this.

l = [1,2,3,[11,22,33,44], 4,5,6, [22,33,44], 5, [99,88]]
k = 0
for k in range(6):
    new_l = []
    for i in range(len(l)):
        if isinstance(l[i], list):
            new_l.append(np.random.choice(l[i]))
        else:
            new_l.append(l[i])
    print(new_l)
    print("\n")

Answer 1

The key function to use is choice from the random module, which randomly selects a value from any iterable object with a known size. All such objects have a __getitem__ method as well as a __len__ method (both of which are needed to apply the choice function), so the builtin function hasattr can be used to check whether choice can be applied or not. The solution becomes straightforward:

from random import choice

l = [1,2,3,[11,22,33,44], 4,5,6, [22,33,44], 5, [99,88]]

for n in range(6):
    print([choice(item) if hasattr(item,'__getitem__') else item for item in l])

Answer 2

You can repeat this procedure 6 times. It randomly reorders your inner lists and takes the first value.

l = [1,2,3,[11,22,33,44], 4,5,6, [22,33,44], 5, [99,88]]

for i in range(6):
    print([np.random.permutation(i)[0] if type(i) == list 
           else np.random.permutation([i])[0] 
           for i in l])

Answer 3

You can try this one too:

import random

l = [1,2,3,[11,22,33,44], 4,5,6, [22,33,44], 5, [99,88]]

res = [random.choice(l[i]) if type(l[i]) is list else l[i] for i in range(len(l))]

print(res)

Possible outputs:

[1, 2, 3, 44, 4, 5, 6, 22, 5, 99]
[1, 2, 3, 11, 4, 5, 6, 22, 5, 99]
[1, 2, 3, 22, 4, 5, 6, 33, 5, 88]