Shuffle a python list without using the built-in function

Question

I\'m working on writing two different shuffle functions.

The first shuffle function must take a list and return a new list with the elements shuffled into a random o

Answer 1

You might find that this implementation for shuffling suits your needs. Make sure that you note the difference between the two functions before using them.

import copy
import random


def main():
    my_list = list(range(10))
    print(my_list)
    print(shuffle(my_list))
    print(my_list)
    shuffle_in_place(my_list)
    print(my_list)


def shuffle(container):
    new_container = copy.copy(container)
    shuffle_in_place(new_container)
    return new_container


def shuffle_in_place(container):
    for index in range(len(container) - 1, 0, -1):
        other = random.randint(0, index)
        if other == index:
            continue
        container[index], container[other] = container[other], container[index]


if __name__ == '__main__':
    main()

Answer 2

You can use this:

import random
def shuffle(lst):
     return sorted(lst ,key = lambda i,random.ramdom())

Answer 3

Plan: Go through the list from the beginning starting with element 0; find a new random position for it, say 6, put 0’s value in 6 and 6’s value in 0. Move on to element 1 and repeat this process, and so on through the rest of the list

import random
iteration = random.randint(2, 100)
temp_var = 0
while iteration > 0:
    # We will be swapping the value of i for j.
    # And then setting j to what i was using the temp_var place holder.
    for i in range(1, len(my_list)): # have to use range with len()
        for j in range(1, len(my_list) - i):
            # Using temp_var as my place holder so I don't lose values
            temp_var = my_list[i]
            my_list[i] = my_list[j]
            my_list[j] = temp_var

        iteration -= 1