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 a seed (448 for example).
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, to return the list in its original order ?
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.
random.Random instance to seed instead of using the default one. This answer made me search for how to do it in a thread-safe manner.A simple check on the python docs http://docs.python.org/library/random.html tells you about
random.seed([x])
which you can use to initialize the seed.
To get the items in the order of your initial again, set the seed again and get the random numbers again. You can then use this index to get the content in the list or just use the index for whatever.
You’d just sort the list and it’d be in sorted order again.
