random.choice() returns same value at the same second, how does one avoid it?

Question

I have been looking at similar questions regarding how to generate random numbers in python. Example: Similar Question - but i do not have the problem that the randomfunctio

Answer 1

You could possibly improve matters by using random.SystemRandom() as follows:

import random

sys_random = random.SystemRandom()

def getRandomID():
    token = ''
    letters = "abcdefghiklmnopqrstuvwwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890"
    for i in range(1, 36):
        token = token + sys_random.choice(letters)
    return token

print(getRandomID())

This attempts to use the os.urandom() function which generates random numbers from sources provided by the operating system. The .choices() function could also be used to return a list of choices in a single call, avoiding the string concatenation:

import random

sys_random = random.SystemRandom()

def getRandomID():
    letters = "abcdefghiklmnopqrstuvwwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890"
    return ''.join(sys_random.choices(letters, k=35))

print(getRandomID())

Answer 2

def getRandomID(n):

    import datetime
    import random

    random.seed(datetime.datetime.now())

    letters = "abcdefghiklmnopqrstuvwwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890"

    idList = [ ''.join([random.choice(letters) for j in range(1,36)]) for i in range(n)]

    return idList

this script in the 3rd test of 10 million ids again have made them all unique

changing for loop to list comprehension did speedup quite a bit.

>>> listt = getRandomID(10000000)
>>> print(len(listt))
10000000

>>> setOfIds = set(listt)
>>> print(len(setOfIds))
10000000

this script uses permutations with repetition: 62 choose 35, to theoretically total number of ids is quite big it is pow(62,35)

541638008296341754635824011376225346986572413939634062667808768

Answer 3

Another option would be to update the seed with the previous result to get a pseudorandom sequence. An option would be old_seed XOR result or just the result.