python shuffle such that position will never repeat

Question

I'd like to do a random shuffle of a list but with one condition: an element can never be in the same original position after the shuffle.

Is there a one line way to do such in python for a list?

Example:

list_ex = [1,2,3]

each of the following shuffled lists should have the same probability of being sampled after the shuffle:

list_ex_shuffled = [2,3,1]
list_ex_shuffled = [3,1,2]

but the permutations [1,2,3], [1,3,2], [2,1,3] and [3,2,1] are not allowed since all of them repeat one of the elements positions.

NOTE: Each element in the list_ex is a unique id. No repetition of the same element is allowed.

Any ideas? thanks!

Answer 1

Randomize in a loop and keep rejecting the results until your condition is satisfied:

import random

def shuffle_list(some_list):
    randomized_list = some_list[:]
    while True:
        random.shuffle(randomized_list)
        for a, b in zip(some_list, randomized_list):
            if a == b:
                break
        else:
            return randomized_list

Answer 2

I'd describe such shuffles as 'permutations with no fixed points'. They're also known as derangements.

The probability that a random permutation is a derangement is approximately 1/e (fun to prove). This is true however long the list. Thus an obvious algorithm to give a random derangement is to shuffle the cards normally, and keep shuffling until you have a derangement. The expected number of necessary shuffles is about 3, and it's rare you'll have to shuffle more than ten times.

(1-1/e)**11 < 1%

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