Python random sequence with seed

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别那么骄傲
别那么骄傲 2020-12-09 10:10

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

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  • 2020-12-09 10:50

    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.

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  • 2020-12-09 10:54
    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.

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