asked 1 year ago
3 Answers
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How can I randomly shuffle a list so that none of the elements remains in its original position?
In other words, given a list A with distinct elements, I'd like to generate a permutation B of it so that
n, a[n] != b[n]
a = [1,2,3,4]
b = [4,1,2,3] # good
b = [4,2,1,3] # good
a = [1,2,3,4]
x = [2,4,3,1] # bad
I don't know the proper term for such a permutation (is it "total"?) thus having a hard time googling. The correct term appears to be "derangement".
After some research I was able to implement the "early refusal" algorithm as described e.g. in this paper [1]. It goes like this:
import random
def random_derangement(n):
while True:
v = [i for i in range(n)]
for j in range(n - 1, -1, -1):
p = random.randint(0, j)
if v[p] == j:
break
else:
v[j], v[p] = v[p], v[j]
else:
if v[0] != 0:
return tuple(v)This should workL
import random
totalrandom = False
array = [1, 2, 3, 4]
it = 0
while totalrandom == False:
it += 1
shuffledArray = sorted(array, key=lambda k: random.random())
total = 0
for i in array:
if array[i-1] != shuffledArray[i-1]: total += 1
if total == 4:
totalrandom = True
if it > 10*len(array):
print("'Total random' shuffle impossible")
exit()
print(shuffledArray)A quick way is to try to shuffle your list until you reach that state. You simply try to shuffle your list until you are left with a list that satisfies your condition.
import random
import copy
def is_derangement(l_original, l_proposal):
return all([l_original[i] != item for i, item in enumerate(l_proposal)])
l_original = [1, 2, 3, 4, 5]
l_proposal = copy.copy(l_original)
while not is_derangement(l_original, l_proposal):
random.shuffle(l_proposal)
print(l_proposal)