Random.Choice from a list that never compares two list items twice (Python)

Question

I'm currently trying to create a program that randomly selects two elements at a time from a list for a user to compare, WITHOUT comparing two elements twice OR the same element for myself .

list1 = ['item1', 'item2', 'item3', 'item4', 'item5'] 

In list1 , I would like to use random.choice:

 item_chosen = random.choice(list1) print(item_chosen) item_chosen2 = random.choice(list1) print(item_chosen2) 

The problem here is that item_chose can be the same as item_chosen2 , and that if I set the loop, then the same comparisons will be performed more than once. In addition, the program must understand when all elements of the list are compared with each other (too much time would be required for manual comparison between all elements in each combination).

Answer 1

Use random.sample :

 item1, item2 = random.sample(list1, 2) 

This gets two unique items from list1 (i.e. item1 and item2 guaranteed to be different).

Answer 2

Shuffle the list (using random.shuffle ) and take items in order from the shuffled list twice.

Answer 3

You should random.shuffle() use random.shuffle() instead of random.choice() :

 list1 = ['item1', 'item2', 'item3', 'item4', 'item5'] random.shuffle(list1) item_chosen = list1.pop() print(item_chosen) item_chosen2 = list1.pop() print(item_chosen2)