How to create a list containing random pairs from an original list?-CodePudding

I have a list:

lst = ['ab', 'cd','ef', 'gh', 'ij', 'mn', 'op', 'qr', 'st', 'uv', 'wx', 'yz']

I would like to take 2 random values from this list and put them in to a new list as pairs until the original list is empty.

For example:

new_list = [('ab', 'ef'), ('ij', 'yz') exc. ] lst = []

How can I do this using a while and for loop?

I've tried using this method to generate a random pair from the list:

random_lst = random.randint(0,len(lst)-1)

However I'm not sure how to remove the values from the original lsit and then add them to the new list as pairs.

CodePudding user response：

I'm sure there are lots of ways. Here's a simple one.

lst = ['ab', 'cd','ef', 'gh', 'ij', 'mn', 'op', 'qr', 'st', 'uv', 'wx', 'yz']
result = []
random.shuffle(lst)
for i in range(0, len(lst), 2):
    result.append((lst[i], lst[i 1]))

CodePudding user response：

Try this

import random

lst = ['ab', 'cd','ef', 'gh', 'ij', 'mn', 'op', 'qr', 'st', 'uv', 'wx', 'yz']
new_list = []

for i in range(len(lst)//2):
    # Get a random index in the current list
    idx = random.randint(0,len(lst)-1)
    # Remove the respective element and store it
    element1 = lst.pop(idx)
    # Get another random index in the current list
    idx = random.randint(0,len(lst)-1)
    # Remove and store that element as well
    element2 = lst.pop(idx)
    # Create an entry (tuple) in the final list
    new_list.append((element1, element2))

print(new_list)

The output for me is [('yz', 'ij'), ('wx', 'ab'), ('st', 'cd'), ('gh', 'uv'), ('qr', 'ef'), ('op', 'mn')]

CodePudding user response：

The sample() function from the random module is ideal for this

from random import sample as SAMPLE

lst = ['ab', 'cd','ef', 'gh', 'ij', 'mn', 'op', 'qr', 'st', 'uv', 'wx', 'yz']
output = []

while lst:
    output.append(SAMPLE(lst, min(2, len(lst))))
    for e in output[-1]:
        lst.remove(e)

print(output)

Note:

min() function is essential for the case where the input list contains an odd number of elements

CodePudding user response：

There are a lot of approaches. With your method, you could try:

import random
lst = ['ab', 'cd','ef', 'gh', 'ij', 'mn', 'op', 'qr', 'st', 'uv', 'wx', 'yz']
random_lst = []
while lst:
  random_lst.append((lst.pop(random.randint(0,len(lst)-1)), lst.pop(random.randint(0,len(lst)-1))))

print(random_lst)

A slight modification to be a bit faster would be:

import random
lst = ['ab', 'cd','ef', 'gh', 'ij', 'mn', 'op', 'qr', 'st', 'uv', 'wx', 'yz', 'xr']
random_lst = []
random.shuffle(lst)
while lst:
  random_lst.append((lst.pop(), lst.pop()))

print(random_lst)

This example accounts for odd length of lists:

import random
lst = ['ab', 'cd','ef', 'gh', 'ij', 'mn', 'op', 'qr', 'st', 'uv', 'wx', 'yz', 'xr']
random_lst = []
random.shuffle(lst)
while lst:
  if len(lst) == 1:
    # The list is not even
    random_lst.append(lst.pop())
    break
  random_lst.append((lst.pop(), lst.pop()))

print(random_lst)

Check out the results: https://colab.research.google.com/drive/19gwP5zPGPXXjGx1AR6VjNkXIIqa16oep#scrollTo=ySBobtc3rtsj

If you want to work with larger lists, I'll suggest checking out numpy.

CodePudding user response：

You can try something using np.random.choice:

new_list = []

while lst:
    tmp=np.random.choice(lst, 2).tolist()
    for t in tmp:
        lst.remove(t)
    new_list.append(tmp)
    
print(new_list)

[['op', 'wx'], ['yz', 'cd'], ['ef', 'qr'], ['mn', 'ij'], ['uv', 'gh'], ['ab', 'st']]