Python random sequence with seed

Question

I'm doing this for a school project (so I can't use any advanced features) and I'm using Python 2.6.6.

I have a list of numbers from 1 to 1000 and my seed will be, lets say, 448.

How can I generate a random sequence with that seed so that the numbers in my list will be in a different index?

And is it possible, knowing the seed, return the elements in my list to the initial position?

Sorry if my question is confusing but English is not my native language.

Thanks.

docs.python.org/release/2.6.6/library/random.html will tell you everything you need to know — Rafe Kettler
I really hope that 448 just happened to be a redicoulsly small number. For your input, there are factorial(1000) sort options, so in order to get proper psaudo-random you will need a seed of this magnitude. — lkanab

Hugh Bothwell Hugh Bothwell · Accepted Answer · 2010-12-29T20:58:58

import random
SEED = 448

myList = [ 'list', 'elements', 'go', 'here' ]
random.seed(SEED)
random.shuffle(myList)

print myList

results in

['here', 'go', 'list', 'elements']

Your list is now pseudorandomized.

'Pseudo' is important, because all lists having the same seed and number of items will return in the same 'random' order. We can use this to un-shuffle your list; if it were truly random, this would be impossible.

Order = list(range(len(myList)))
# Order is a list having the same number of items as myList,
# where each position's value equals its index

random.seed(SEED)
random.shuffle(Order)
# Order is now shuffled in the same order as myList;
# so each position's value equals its original index

originalList = [0]*len(myList)   # empty list, but the right length
for index,originalIndex in enumerate(Order):
    originalList[originalIndex] = myList[index]
    # copy each item back to its original index

print originalList

results in

['list', 'elements', 'go', 'here']

Tada! originalList is now the original ordering of myList.

Python random sequence with seed

2 Answers