算法从列表中访问两个随机元素，将它们存储在变量中并从 Python 3.x 中的原始列表中删除元素？

Question

I have a list, that so far has 4 elements. I can use the random.sample command to pick 2 elements at random

teamslist = "apple" , "banana" , "orange", "clementines"
teamrandom = random.sample(teamslist, k = 2)

Is there anyway for the randomly picked elements stored in teamrandom, to be removed from list "teamlist", so that only the remaining 2 strings are left in the list?

This way, when the command is run again, it only picks the 2 remaining elements from the list?

No element should be chosen more than once

Answer 1

You can simply use random.shuffle to shuffle the list(change the order of its elements) and then pop an element from it each time you need a random one:

import random
teamslist = ["apple" , "banana" , "orange", "clementines"]
random.shuffle(teamslist)

and when you want a random element

element = teamslist.pop()

Note: list.pop() returns the last element after removing it from a list

EDIT:

As stated in the comments, this will lose the order in the original list. So a simple fix would be:

import random
teamslist = ["apple" , "banana" , "orange", "clementines"]
shuffled = teamslist.copy()
random.shuffle(shuffled)

and then to take a new element, just use the same syntax:

element = shuffled.pop()

Answer 2

pop() a random index in the list and store in a new list:

import random

teamslist = ["apple" , "banana" , "orange", "clementines"]

teamsrandom = [teamslist.pop(random.randint(0, len(teamslist) - 1)) for _ in range(2)]

print(teamsrandom)
print(teamslist)

Output:

['clementines', 'orange']
['apple', 'banana']

Answer 3

You can use randint and pop :

from random import randint
teamslist = ["apple" , "banana" , "orange", "clementines"]
teamrandom = [teamslist.pop(randint(0,len(teamslist)-1)),
              teamslist.pop(randint(0,len(teamslist)-1))]

print(teamslist)
print(teamrandom)

Output:

['banana', 'clementines']
['apple', 'orange']