如何在Python中将列表分为两个唯一列表？

Question

Hi I have a list as following: listt = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o'] 15 members.

I want to turn it into 3 lists, I used this code it worked but I want unique lists. this give me 3 lists that have mutual members.

import random

listt = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o']
print(random.sample(listt,5))
print(random.sample(listt,5))
print(random.sample(listt,5))

Answer 1

Try this:

from random import shuffle


def randomise():
    listt = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o']
    shuffle(listt)
    return listt[:5], listt[5:10], listt[10:]

print(randomise())

This will print (for example, since it is random):

(['i', 'k', 'c', 'b', 'a'], ['d', 'j', 'h', 'n', 'f'], ['e', 'l', 'o', 'g', 'm'])

Answer 2

Use [:] for slicing all members out of the list which basically copies everything into a new object. Alternatively just use list(<list>) which copies too:

print(random.sample(listt[:],5))

In case you want to shuffle only once, store the shuffle result into a variable and copy later:

output = random.sample(listt,5)
first = output[:]
second = output[:]
print(first is second, first is output)  # False, False

and then the original list can be modified without the first or second being modified.

For nested lists you might want to use copy.deepcopy() .

Answer 3

Instead of sampling your list three times, which will always give you three independent results where individual members may be selected for more than a single list, you could just shuffle the list once and then split it in three parts. That way, you get three random subsets that will not share any items:

>>> random.shuffle(listt)
>>> list[0:5]
>>> listt[0:5]
['b', 'a', 'f', 'e', 'h']
>>> listt[5:10]
['c', 'm', 'g', 'j', 'o']
>>> listt[10:15]
['d', 'l', 'i', 'n', 'k']

Note that random.shuffle will shuffle the list in place, so the original list is modified. If you don't want to modify the original list, you should make a copy first.

If your list is larger than the desired result set, then of course you can also sample your list once with the combined result size and then split the result accordingly:

>>> sample = random.sample(listt, 5 * 3)
>>> sample[0:5]
['h', 'm', 'i', 'k', 'd']
>>> sample[5:10]
['a', 'b', 'o', 'j', 'n']
>>> sample[10:15]
['c', 'l', 'f', 'e', 'g']

This solution will also avoid modifying the original list, so you will not need a copy if you want to keep it as it is.

Answer 4

If it doesn't matter to you which items go in each list, then you're better off partitioning the list into thirds:

In [23]: L =  ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o']                                                                                                     

In [24]: size = len(L)                                                                                                                                                          

In [25]: L[:size//3]                                                                                                                                                            
Out[25]: ['a', 'b', 'c', 'd', 'e']

In [26]: L[size//3:2*size//3]                                                                                                                                                   
Out[26]: ['f', 'g', 'h', 'i', 'j']

In [27]: L[2*size//3:]                                                                                                                                                          
Out[27]: ['k', 'l', 'm', 'n', 'o']

If you want them to have random elements from the original list, you'll just need to shuffle the input first:

random.shuffle(L)